# Precalculus Mathematics 2 Study Guide

When it comes to college courses, Precalculus Mathematics 2 is a course that you will have to understand how to do the following things to get ready for the actual final exam of this respective mathematics course:

- Using half-angle formulas
- Using summation formulas

- Solving word problems
- Converting polar coordinates to rectangular coordinates
- Finding values of trigonometric functions
- Converting rectangular coordinates to polar coordinates
- Determining one's amplitude and period (without graphing)
- Finding the exact value of expressions
- Both with and without a calculator
- Using trivial identities

- Finding the area of triangles
- Converting rectangular equations to polar equations
- Using product-sum identities
- Using sum-product identities
- Finding sides and angles of a triangle
- Using definitions and identities
- Naming quadrants
- Solving triangles
- Solving trigonometric equations on an interval
- Establishing identities
- Simplifying trigonometric expressions
- Finding equations for a graph

P.S. I am writing this study guide based on Barry University's MAT 110 final exam study guide that I was given. However, if you would like to see the full document of Barry University's MAT 110 final exam study guide, then you can check my version of it on either Academia, CourseHero, or even on my LinkedIn. Click here if you want the Precalculus Mathematics 1 Study Guide.

#### Contents

- Section \(A_1\). - Half-Angle Formulas
- Section \(A_2\). - Summation Identities
- Section B. - Solving Word Problems
- Section C. - Polar to Rectangular Coordinates
- Section D. - Double-Angle Formulas
- Section E. - Finding Amplitude and Period of Sinusoidal Functions
- Section F. - Finding Exact Values of Expressions
- Section G. - Finding the Area of a Triangle
- Section H. - Rectangular to Polar Equations
- Section I. - Polar to Rectangular Equations
- Section J. - Product-Sum Formulas
- Section K. - Sum-Product Formulas
- Section L. - Find the Indicated Trigonometric Functions of the Given Angle
- Section M. - Find the Exact Value of a Trigonometric Function
- Section N. - Find the Exact Value of an Expression
- Section O. - Naming the Quadrant
- Section P. - Solving a Triangle
- Section Q. - Solving a (Possible) Triangle
- Section R. - Finding the Exact Value of the Indicated Trigonometric Function
- Section S. - Find the Exact Value of an Expression - Part 2
- Section T. - Find the Exact Value of an Expression - Part 3
- Section U. - Using Definitions and Identities
- Section V. - Solving Equations on an Interval
- Section W. - Establish the Identity
- Section X. - Simplifying the Trigonometric Expressions
- Section Y. - Solving the Equation on an Interval
- Section Z. - Find the Equation for the Graph
- Section AA. - Finding Exact Values of Expressions

## Section \(A_1\). - Half-Angle Formulas

When it comes to using the half-angle formulas, you have to understand what the half-angle formulas are, which can be expressed as the following formulas: \[\begin{matrix} & \textup{\textbf{Half-Angle Formulas}}& \\ \textup{Common Formulas} & &\textup{Uncommon Formulas} \\\\ \sin \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1-\cos \left(x\right)}{2}} & &\csc \left(\frac{x}{2}\right)=\frac{1}{\sin\left(\frac{\theta }{2}\right)}=\pm \sqrt{\frac{2}{1-\cos \left(x\right)}}\\\\ \cos \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1+\cos \left(x\right)}{2}} & &\sec\left(\frac{x}{2}\right)=\frac{1}{\cot\left(\frac{\theta }{2}\right)}=\pm \sqrt{\frac{2}{1+\cos \left(x\right)}}\\\\ \tan \left(\frac{\theta }{2}\right)=\frac{\sin\left(\frac{\theta }{2}\right)}{\cos\left(\frac{\theta }{2}\right)}=\pm \sqrt{\frac{1-\cos \left(x\right)}{1+\cos \left(x\right)}} & &\cot\left(\frac{x}{2}\right)=\frac{1}{\tan\left(\frac{\theta }{2}\right)}=\pm \sqrt{\frac{1+\cos \left(x\right)}{1-\cos \left(x\right)}} \end{matrix}\]

Before you use a half-angle formula, make sure that you have the degrees as a fraction with \(2\) being its denominator hence why the half-angle formulas deal with \(\frac{\theta}{2}\). It's common to use both a half-angle formula along with a trivial identity.

Use the half-angle formulas to find the exact value of the following trigonometric function: \[\sin\:22.5^{\circ}\]

Since you are looking for sine at \(22.5^{\circ}\), then this means it will be lying in the 1st quadrant, which means that the final answer will be positive. So, if \(22.5\div \frac{1}{2}=22.5\cdot 2=45\), then \[\sin\:22.5^{\circ}=\sin \left(\frac{45^{\circ}}{2}\right).\] Then, you can use sine's half angle identity, which is \[\sin \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1-\cos \left(x\right)}{2}}\] in order to go to the next step. This means the following: \[\sin \left(22.5^{\circ}\right)=\sin \left(\frac{45^{\circ}}{2}\right)=\sqrt{\frac{1-\cos \left(45^{\circ}\right)}{2}}\]From there, you would have to use one of cosine's trivial identity, which is basically \(\cos \left(45^{\circ}\right)\) in this case, which means that \(\cos \left(45^{\circ }\right)=\frac{\sqrt{2}}{2}\). From there, you can simplify everything from there, which means the following: \[\begin{matrix} \sqrt{\frac{1-\cos \left(45^{\circ }\right)}{2}}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{\frac{2}{2}-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}=\boxed{\mathbf{{\color{red} \frac{1}{2}\sqrt{2-\sqrt{2}} }}} \end{matrix}\]

## Section \(A_2\). - Summation Identities

Depending on the professor, some will teach summation identities while teaching half-angle formulas as an alternative way to solve half-angle formulas or solve as an addition to half-angle formulas. So, here are the 6 summation identities (2 for sine, 2 for cosine, and 2 for tangent): \[\begin{matrix} \textbf{\textbf{Summation Formulas}}\\ \textup{Sine}\\ \sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\cos\left(a\right)\sin\left(b\right)\\ \sin\left(a-b\right)=\sin\left(a\right)\cos\left(b\right)-\cos\left(a\right)\sin\left(b\right)\\\\ \textup{Cosine}\\ \cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)\\ \cos\left(a-b\right)=\cos\left(a\right)\cos\left(b\right)+\sin\left(a\right)\sin\left(b\right)\\\\ \textup{Tangent}\\ \begin{matrix} \tan\left(a+b\right)=\frac{\tan\left(a\right)+\tan\left(b\right)}{1-\tan\left(a\right)\tan\left(b\right)}&\tan\left(a-b\right)=\frac{\tan\left(a\right)-\tan\left(b\right)}{1+\tan\left(a\right)\tan\left(b\right)} \end{matrix} \end{matrix}\]

Use the half-angle formulas to find the exact value of the following trigonometric function: \[\cos\:165^{\circ}\]

Since you are looking for cosine at \(165^{\circ}\), then this means it will be lying in the 4th quadrant, which means that the final answer will be negative. So, if \(165\div \frac{1}{2}=165\cdot 2=330\), then \[\cos\:165^{\circ}=\cos \left(\frac{330^{\circ}}{2}\right).\] Then, you can use cosine's half angle identity, which is \[\cos \left(\frac{x}{2}\right)=\pm \sqrt{\frac{1+\cos \left(x\right)}{2}}\] in order to go to the next step. This means the following: \[\cos \left(165^{\circ}\right)=\cos \left(\frac{330^{\circ}}{2}\right)=-\sqrt{\frac{1+\cos \left(330^{\circ}\right)}{2}}\]Instead of using the trivial identity of \(\cos \left(330^{\circ}\right)\), we will be using the following identity in order to make it easier for us: \[\cos \left(x\right)=\sin \left(90^{\circ }-x\right),\] which means the following: \[-\sqrt{\frac{1+\cos \left(330^{\circ }\right)}{2}}=-\sqrt{\frac{1+\sin \left(90^{\circ }-330^{\circ }\right)}{2}}=-\sqrt{\frac{1+\sin \left(-240^{\circ }\right)}{2}}\]

Now, recall that \(\sin \left(-x\right)=-\sin \left(x\right)\), which means that \[-\sqrt{\frac{1+\sin \left(-240^{\circ }\right)}{2}}=-\sqrt{\frac{1-\sin \left(240^{\circ }\right)}{2}}\]

From there, you would have to use one of sine's trivial identity, which is basically \(\sin \left(240^{\circ}\right)\) in this case, which means that \(\sin \left(240^{\circ }\right)=-\frac{\sqrt{3}}{2}\). From there, you can simplify everything from there, which means the following: \[\begin{matrix} \sqrt{\frac{1-\sin \left(240^{\circ }\right)}{2}}\\ =-\sqrt{\frac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}}\\ =-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{\frac{2}{2}+\frac{\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{2+\sqrt{3}}{4}}\\ =-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\\ =-\frac{\sqrt{2+\sqrt{3}}}{2}\\ =\boxed{\mathbf{{\color{red} -\frac{1}{2}\sqrt{2+\sqrt{3}} }}} \end{matrix}\]

Use the summation formulas to find the exact value of the following trigonometric function: \[\cos\:165^{\circ}\]

So, when it comes to summation formulas, the first thing to do it ironically "simplifying" \(\cos\:165^{\circ}\) by using the following identity: \[\cos \left(x\right)=\sin \left(90^{\circ }-x\right).\] Therefore, if \(\cos\:165^{\circ}\) and \(\cos \left(x\right)=\sin \left(90^{\circ }-x\right)\), then this would mean the following: \[\begin{matrix} \cos \left(165^{\circ }\right)=\sin \left(90^{\circ }-165^{\circ }\right)=\sin \left(-75^{\circ }\right) \end{matrix}\]Next, you have to recall the trigonometric property of \(\sin \left(-x\right)=-\sin \left(x\right)\), which means that \[\sin \left(-75^{\circ }\right)=-\sin \left(75^{\circ }\right)\]

Now, you can actually just let \[-\sin \left(75^{\circ }\right)=-\sin \left(30^{\circ }+45^{\circ }\right)\] Since that is the case, then this is when you now can use sine's positive summation identity, which is the following: \[\sin \left(x+y\right)=\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right).\]

Afterwards, you can just substitute then use trivial identities then simplifying everything to find the exact value of \(\cos\:165^{\circ}\), which can be done by following the following: \[\begin{matrix} \sin \left(x+y\right)&=\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)\\ -\sin \left(x+y\right)&=-[\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)]\\ -\sin \left(30^{\circ }+45^{\circ }\right)&=-[\sin \left(30^{\circ }\right)\cos \left(45^{\circ }\right)+\cos \left(30^{\circ }\right)\sin \left(45^{\circ }\right)]\\ &=-\left [\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\right ]\\ &=-\left [\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right ]\\ &=-\left [\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}\right ]\\ &=-\left [\frac{1+\sqrt{3}}{2\sqrt{2}}\right ]\\ &=-\left [ \frac{\left(1+\sqrt{3}\right)\cdot \sqrt{2}}{2\sqrt{2}\cdot \sqrt{2}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{2\sqrt{4}}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{2(2)}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{2} \right ]\\ &=-\left [ \frac{1}{2}\sqrt{2+\sqrt{3}} \right ]\\ &=\boxed{\mathbf{{\color{red} -\frac{1}{2}\sqrt{2+\sqrt{3}} }}} \end{matrix}\]

## Section B. - Solving Word Problems

In precalculus, the majority of word problems are ironically going to be triangle-based, but more specifically right-angled triangles. Since that is the case, then it is commonly best to draw a triangle just so you know what you're dealing with. If they give you a triangle, then there is no need to draw anything, but just label anything if needed. In a right-angled triangle, the right angle will always be \(\measuredangle C\).

When it comes to solving, you can use either to Soc-Cah-Toa method or even using the law of sines, which both can be noted as the following: \[\begin{matrix} \textup{\textbf{Soc-Cah-Toa}}&\textbf{\textbf{Law of Sines}}\\ \sin(\theta)=\frac{\textup{opposite}}{\textup{hypotenuse}}=\frac{\textup{side}\:a}{\textup{side}\:c}&\frac{a}{\sin\:A}=\frac{b}{\sin\:B}=\frac{c}{\sin\:C}\\\\ \cos(\theta)=\frac{\textup{adjacent}}{\textup{hypotenuse}}=\frac{\textup{side}\:b}{\textup{side}\:c}&\\\\ \tan(\theta)=\frac{\textup{opposite}}{\textup{adjacent}}=\frac{\textup{side}\:a}{\textup{side}\:b}& \end{matrix}\]

A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake \(150\) feet from a piling that is directly across from a pier on the other side of the lake. From his transit, the angle between the piling and the pier is \(30^{\circ}\) degrees. What is the distance between the piling and the pier to the nearest foot? \[\boxed{\textup{P.S. You would have to use Soh-Cah-Toa in order to solve this problem.}}\]

When it comes to the right triangle for this problem, the triangle will be in the form of Pier-Piling-Transit. The adjacent side, also known as \(\textup{side\:a}\), which is the distance between the piling to the transit, is \(\textup{150\:ft}\). The opposite side, also known as \(\textup{side\:b}\), which is the distance between the pier to the piling, is \(\textup{x}\). The hypotenuse, also known as \(\textup{side\:c}\), which is the distance between the transit to the pier, is not going to be used for solving this problem. However, the angle of the pier, also known as \(\theta\), is \(30^{\circ}\). So, in terms of drawing out this triangle, you can format it like the following:Now, when it comes to this problem, this means that you are looking for the distance between the pier to the piling, which means doing the following: \[\begin{matrix} \tan \theta =\frac{\textup{opposite}}{\textup{adjacent}}\\\\ \tan 30^{\circ} =\frac{x}{150} \end{matrix}\]

Now, in order to solve this, you have to solve for \(x\), which means doing the following: \[\begin{matrix} \tan 30^{\circ} =\frac{x}{150}\\ 150\tan 30^{\circ} =x\\ 150\left ( \frac{1}{\sqrt{3}} \right ) =x\\ \frac{150}{\sqrt{3}} =x\\ \frac{150}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} =x\\ \frac{150\sqrt{3}}{3} =x\\ \frac{50\sqrt{3}}{1} =x\\ 50\sqrt{3} =x\\ 86.60254 \approx x\\ 87 \approx x\\ \boxed{\mathbf{{\color{red} x\approx 87\: feet }}} \end{matrix}\]

A radio transmission tower is \(190\) feet tall. How long should a guy wire be if it is to be attached \(15\) feet from the top and is to make an angle of \(30^{\circ}\) with the ground? Give your answer to the nearest tenth of a foot. \[\boxed{\textup{P.S. You would have to use the law of sines in order to solve this problem.}}\]

Since the guy wire is supposed to be "attached \(15\) feet from the top" from the given \(190\) foot tower, which means that you have to do some subtraction in order to properly find \(\text{side a}\), which can be done by doing the following:This means that \(\text{side a}\) is going to be \(\text{175 feet}\). So, in terms of drawing out this triangle, you can format it like the following:

So, you are given \(\text{side a}\) along with \(\text{angles a and c}\) (with \(\text{angle c}\) being the \(90^{\circ}\) angle), which means that you need to look for \(\text{side c}\), which means that you would have to do the following in order to find the length of the guy wire: \[\begin{matrix} \frac{a}{\sin A}=\frac{c}{\sin C}\\\\ \frac{175}{\sin 30^{\circ}}=\frac{x}{\sin 90^{\circ}}\\ 175\sin 90^{\circ}=x\sin 30^{\circ}\\ 175(1)=x\left ( \frac{1}{2} \right )\\ 175=\frac{x}{2}\\ 175\cdot 2=\frac{x}{2}\cdot 2\\ 350=x\\ \boxed{\mathbf{{\color{red} x=350\: feet }}} \end{matrix}\]

## Section C. - Polar to Rectangular Coordinates

When it comes to converting polar coordinates to rectangular coordinates, the main thing to know is understand the difference between the polar and rectangular coordinates, which can be noted by the following: \[\begin{matrix} \textup{\textbf{Polar Coordinates}}&\textbf{\textbf{Rectangular Coordinates}}\\ (r,\theta)&(x,y) \end{matrix}\]

\(r\) is the distance from the origin while \(\theta\) is the angle that's measured from the \(x\)-axis to the point. However, when we need to convert polar coordinates to rectangular coordinates, you would have to do the following: \[\begin{matrix} \textup{\textbf{If}}&\textbf{\textbf{Then}}\\ x=r \cos\: \theta&\\ \textup{and}&(x,y) =(r \cos\: \theta,r \sin\: \theta)\\ y=r \sin\: \theta&\\ \end{matrix}\]

Now, you will be using trivial identities to simplify the coordinates.

The polar coordinates of a point are given. Find the rectangular coordinates of the following point: \[\left(7,\frac{2\pi }{3}\right)\]

So, if \(\left(7,\:\frac{2\pi }{3}\right)=\left(x,y\right)\), then this means that \(x=7=r\) and \(y=\frac{2\pi }{3}=\theta\). This means the following: \[\begin{matrix} \left(7,\frac{2\pi }{3}\right)\\ (x,y)\\ (r,\theta)\\\\ (r \cos \theta,r \sin \theta)\\ (7 \cos \frac{2\pi }{3},7 \sin \frac{2\pi }{3})\\ (7 \left ( -\frac{1}{2} \right ),7 \left ( \frac{\sqrt{3}}{2} \right ))\\ \boxed{\mathbf{{\color{red} \left(-\frac{7}{2},\frac{7\sqrt{3}}{2} \right) }}} \end{matrix}\]

## Section D. - Double-Angle Formulas

When it comes to finding the exact value of a trigonometric function by using a double-angle formula, you have to understand what the double-angle formulas are, which can be expressed as the following formulas: \[\begin{matrix} & \textup{\textbf{Double-Angle Formulas}}& \\ \textup{Common Formulas} & &\textup{Uncommon Formulas} \\\\ \sin\left(2\theta \right)=2\:\sin\theta\: \cos\theta & &\sin\left(2\theta \right)=\frac{1}{\sin\left(2\theta \right)}=\frac{1}{2\:\sin\theta\: \cos\theta} \\\\ \begin{matrix} \cos\left(2\theta \right)&=\cos^2\theta-\sin^2\theta\\ &=2\cos^2\theta-1\\ &=1-2\sin^2\theta \end{matrix} & & \begin{matrix} \sec\left(2\theta \right)&=\frac{1}{cos\left(2\theta \right)}&=\frac{1}{\cos^2\theta-\sin^2\theta}\\ &&=\frac{1}{2\cos^2\theta-1}\\ &&=\frac{1}{1-2\sin^2\theta} \end{matrix}\\\\ \tan(2\theta)=\frac{2\tan\theta}{1-tan^2\theta} & &\cot(2\theta)=\frac{1}{\tan(2\theta)}=\frac{1-tan^2\theta}{2\tan\theta}\\\\ \end{matrix}\]

Sometimes, you may use only a double-angle formula or even use Soc-Cah-Toa, the Pythagorean Theorem, and a double-angle formula, but it depends on the problem that you are given. Either way, a trivial identity will be used. Also, remember that \(\textup{angle}\:\theta=0\le \theta \le 2\pi\).

Use the information given about the angle \(\theta\), \(0\le \theta \le 2\pi\), to find the exact value of the following indicated trigonometric function: \[\begin{matrix} \sin(\theta)=\frac{8}{17},&0\le \theta \le \frac{\pi }{2}&&&\textup{Find} \cos(2\theta). \end{matrix}\]

\[\boxed{\textup{P.S. You would have to use only a double angle in order to solve this problem.}}\]

If \(\cos(2\theta)=1-2\sin^2\theta\) and \(\sin(\theta)=\frac{8}{17}\), then this means that \(\sin^2(\theta)=(\frac{8}{17})^2\). With that being said, the following is how you can solve this problem: \[\begin{matrix} \cos(2\theta)&=1-2\sin^2\theta\\ &=1-2(\frac{8}{17})^2\\\\ &=1-2(\frac{64}{289})\\\\ &=1-\frac{128}{289}\\\\ &=\frac{289}{289}-\frac{128}{289}\\\\ &=\frac{289-128}{289}\\\\ &=\boxed{\mathbf{{\color{red} \frac{161}{289} }}} \end{matrix}\]

Use the information given about the angle \(\theta\), \(0\le \theta \le 2\pi\), to find the exact value of the following indicated trigonometric function: \[\begin{matrix} \tan(\theta)=\frac{12}{5},&\pi\le \theta \le \frac{3\pi }{2}&&&\textup{Find} \sin(2\theta). \end{matrix}\]

\[\boxed{\textup{P.S. You would have to use both the Soh-Cah-Toa and the Pythagorean Theorem in order to solve this problem.}}\]

Since \(\tan(\theta)=\frac{12}{5}=\frac{\textup{opposite}}{\textup{adjacent}}\), which means you do not have the hypotenuse, which means you need to use the Pythagorean Theorem in order to solve, which means doing the following: \[\begin{matrix} \sqrt{a^2+b^2}=c\\ \sqrt{a^2+o^2}=h\\ \sqrt{(5)^2+(12)^2}=h\\ \sqrt{25+144}=h\\ \sqrt{169}=h\\ \pm 13=h\\ 13=h\\ h=13\\ \end{matrix}\]So, now we know that \(\left\{\begin{matrix} a&=&5\\o&=&12\\h&=&13 \end{matrix}\right.\). With that being that, you need to know sine's double-angle formula, which is \[\sin\left(2\theta \right)=2\sin\theta \cos\theta \] while recall that both \(\sin(\theta)=\frac{o}{h}\) and \(\cos(\theta)=\frac{a}{h}\), which means you will have to do the following in order to find the exact value of this indicated trigonometric function: \[\begin{matrix} \sin(2\theta)&=2\sin\theta \cos\theta\\ &=2(\frac{o}{h}) (\frac{a}{h})\\\\ &=2(\frac{12}{13}) (\frac{5}{13})\\\\ &=2(\frac{60}{169})\\\\ &=\boxed{\mathbf{{\color{red} \frac{120}{169} }}} \end{matrix}\]

## Section E. - Finding Amplitude and Period of Sinusoidal Functions

When it comes to finding the amplitude and/or period of a function. it is best to first know the general sinusoidal function, which is \[\begin{matrix}f(x)=a\sin(bx-c)+d&\textup{while}&b>0\end{matrix}.\] The following formulas can be used when it comes to finding various parts of a sinusoidal function: \[\begin{matrix} \textup{\textbf{Amplitude}}&\textup{\textbf{Period}}&\textup{\textbf{Phase/Horizontal Shift}}&\textup{\textbf{Vertical Shift}}& \\ \left | a \right |&\frac{2\pi}{b}&c&d& \\ \end{matrix}\]

Without graphing the function, determine its amplitude or period as requested. \[\begin{matrix} y=-3\sin x&&\textup{Find the amplitude.} \end{matrix}\]

So, when it comes to finding the amplitude of this function, you can use the general sinusoidal function and see how the given function manages, then this would be expressed as the following: \[\begin{matrix} f(x)&=&a&\sin(bx&-&c)&+d\\ y&=&-3&\sin(1x&-&0)&+0 \end{matrix}\]With that being said, since \(a=-3\), the amplitude is \(\left | a \right |\), which means \[\left | a \right |=\left | -3 \right |=\boxed{\mathbf{{\color{red} 3 }}}\]

Without graphing the function, determine its amplitude or period as requested. \[\begin{matrix} y=\sin 5x&&\textup{Find the period.} \end{matrix}\]

So, when it comes to finding the period of this function, you can use the general sinusoidal function and see how the given function manages, then this would be expressed as the following: \[\begin{matrix} f(x)&=&a&\sin(bx&-&c)&+d\\ y&=&1&\sin(5x&-&0)&+0 \end{matrix}\]With that being said, since \(a=-3\), the period is \(\frac{2\pi}{b}\), which means \[\frac{2\pi}{b}=\boxed{\mathbf{{\color{red} \frac{2\pi}{5} }}}\]

## Section F. - Finding Exact Values of Expressions

When it comes to finding the exact value of trigonometric expressions, trivial identities will be used in order to simplify the given expression.

Find the exact value of the following expression. Do not use a calculator. \[\cos 60^{\circ}+\tan 60^{\circ}\]

\[\begin{matrix} \cos\:60^{\circ }+\tan\:60^{\circ }\\ =\frac{1}{2}+\sqrt{3}\\ =\frac{1}{2}+\frac{2\sqrt{3}}{2}\\ =\boxed{\mathbf{{\color{red} \frac{1+2\sqrt{3}}{2} }}} \end{matrix}\]

## Section G. - Finding the Area of a Triangle

When it comes to finding the area of a triangle, you are then commonly given 2 sides and a drink...I mean 2 sides and an angle. Anyways, based on the given info, you'll always have to half the 2 given sides then multiply them by the sine of the given degree. Either way, the general formula will still be the same format, which will be the following: \[A=\frac{1}{2}(\textup{1st given side})(\textup{2nd given side})\sin(\textup{given angle})\]

To be more specific, here are the following formulas based on which sides and angle that you are given: \[\begin{matrix} \textup{\textbf{Given angle A}}&\textup{\textbf{Given side a}}&\textup{\textbf{Given side a}} \\ \textup{\textbf{Given side b}}&\textup{\textbf{Given angle B}}&\textup{\textbf{Given side b}} \\ \textup{\textbf{Given side c}}&\textup{\textbf{Given side c}}&\textup{\textbf{Given angle C}} \\\\ A=\frac{1}{2}bc\:\sin\:A&A=\frac{1}{2}ac\:\sin\:B&A=\frac{1}{2}ab\:\sin\:C \\ \end{matrix}\]

Also, if there are no given measurements, then write the final answer in "units squared", or \(u^2\), since you are dealing with the "area" of a triangle.

Find the area of the triangle. If necessary, round the answer to two decimal places.

So, from this triangle, you are given \textup{sides a and b} along with \textup{angle C}, which means that you will be using the following version of the formula in order to solve the problem: \[A=\frac{1}{2}ab\:\sin\:C\]With that being said, you can do the following to find the area of this triangle: \[\begin{matrix} A&=\frac{1}{2}ab\sin C\\ &=\frac{1}{2}(6)(8)\sin (110^{\circ})\\ &=\frac{1}{2}(48)\sin (110^{\circ})\\ &=24\sin (110^{\circ})\\ &\boxed{\mathbf{{\color{red} \approx 22.55\:u^2 }}} \end{matrix}\]

## Section H. - Rectangular to Polar Equations

If you remember Section C, you are doing the same thing, but the only difference is that in Section H, you are converting equations instead of converting coordinates. It is the same concept when it comes to using the coordinates in order to convert the equations, which can be re-noted by the following: \[\begin{matrix} \textup{\textbf{Polar Coordinates}}&\textbf{\textbf{Rectangular Coordinates}}\\ (r,\theta)&(x,y) \end{matrix}\]

Like Section C, you have to keep in mind that \(x=r \cos\: \theta\) and \(y=r \sin\: \theta\) along with the following if-then scenario just in case when it comes to converting equations: \[\begin{matrix} \textup{\textbf{If}}&\textbf{\textbf{Then}}\\ r=\sqrt{x^2+y^2}&r^2=x^2+y^2 \end{matrix}\]

The letters \(x\) and \(y\) represent rectangular coordinates. Write the following equation using polar coordinates \((r,\theta)\): \[x^2+y^2-4x=0\]

\[\begin{matrix} x^2+y^2-4x=0\\ (x^2+y^2)-4(x)=0\\ (r^2)-4(r \cos\theta)=0\\ r^2-4r \cos\theta=0\\ r^2=4r \cos\theta\\ \frac{r^2}{r}=\frac{4r \cos\theta}{r}\\ \boxed{\mathbf{{\color{red} r=4 \cos\theta }}} \end{matrix}\]

## Section I. - Polar to Rectangular Equations

If you remember Section H, you are doing the same thing, but the only difference is that in Section I, you are converting a polar equation to a rectangular equation instead of converting a rectangular equation to a polar equation. It is the same concept when it comes to using the coordinates in order to convert the equations, which can be re-noted by the following: \[\begin{matrix} \textup{\textbf{Polar Coordinates}}&\textbf{\textbf{Rectangular Coordinates}}\\ (r,\theta)&(x,y) \end{matrix}\]

Like Section C, you have to keep in mind that \(x=r \cos\: \theta\) and \(y=r \sin\: \theta\) along with the following if-then scenario just in case when it comes to converting equations: \[\begin{matrix} \textup{\textbf{If}}&\textbf{\textbf{Then}}\\ r=\sqrt{x^2+y^2}&r^2=x^2+y^2 \end{matrix}\]

The letters \(r\) and \(\theta\) represent polar coordinates. Write the following equation using rectangular coordinates \((x,y)\): \[r=\frac{5}{1+\cos \left(\theta\right)}\]

\[\begin{matrix} r=\frac{5}{1+\cos\theta}\\ r(1+\cos\theta)=5\\ r+r\cos\theta=5\\ r+(r\cos\theta)=5\\ r+(x)=5\\ r+x=5\\ r=5-x\\ (r)^2=(5-x)^2\\ r^2=x^2-10x+25\\ (r^2)=x^2-10x+25\\ (x^2+y^2)=x^2-10x+25\\ x^2+y^2=x^2-10x+25\\ y^2=-10x+25\\ \boxed{\mathbf{{\color{red} y^2=25-10x }}} \end{matrix}\]

## Section J. - Product-Sum Formulas

Some call these "product-to-sum formulas" while others call them "product-to-sum formulas", but they are basically the same thing. either way, when it comes to expressing a product as a sum while containing only sines and/or cosines, there are 4 possible formulas to use based on the problem that you are given, which can be expressed by the following: \[\begin{matrix} \textup{\textbf{Sine Intro}}&\textbf{\textbf{Cosine Intro}}\\ \sin\left(a\right)\sin\left(b\right)=\frac{1}{2}\left[\cos\left(a-b\right)-\cos\left(a+b\right)\right]&\cos\left(a\right)\cos\left(b\right)=\frac{1}{2}\left[\cos\left(a+b\right)+\cos\left(a-b\right)\right]\\\\

\sin\left(a\right)\cos\left(b\right)=\frac{1}{2}\left[\sin\left(a+b\right)+\sin\left(a-b\right)\right]&\cos\left(a\right)\sin\left(b\right)=\frac{1}{2}\left[\sin\left(a+b\right)-\sin\left(a-b\right)\right]\\\end{matrix}\]

Express the following product as a sum containing only sines or cosines: \[\sin(8\theta)\cos(4\theta)\]

Since it's sine being multiplied by cosine, this means that the you are going to be using \[\sin\left(a\right)\cos\left(b\right)=\frac{1}{2}\left[\sin\left(a+b\right)+\sin\left(a-b\right)\right]\] in order to express the product as a sum.So, with that being said, here is how to express the given product: \[\begin{matrix} \sin\left(a\right)\cos\left(b\right)&=\frac{1}{2}\left[\sin\left(a+b\right)+\sin\left(a-b\right)\right]\\ \sin\left(8\theta \right)\cos\left(4\theta \right)&=\frac{1}{2}\left[\sin\left(8\theta +4\theta \right)+\sin\left(8\theta -4\theta \right)\right]\\ &=\boxed{\mathbf{{\color{red} \frac{1}{2}\left[\sin\left(12\theta \right)+\sin\left(4\theta \right)\right] }}} \end{matrix}\]

## Section K. - Sum-Product Formulas

Some call these "sum-to-product formulas" while others call them "sum-product formulas" with some even calling these "superposition relationships"., but they are basically the same thing. Either way, when it comes to expressing a sum as a product while containing only sines and/or cosines, there are 4 possible formulas (2 are for sums and 2 are for differences) to use based on the problem that you are given, which can be expressed by the following: \[\begin{matrix} \textup{\textbf{\textit{Sums}}}&\textbf{\textbf{\textit{Differences}}}\\ \textup{\textbf{Sine Sum}}&\textbf{\textbf{Sine Difference}}\\ \sin\left(a\right)+\sin\left(b\right)=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)

&\sin\left(a\right)-\sin\left(b\right)=2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\\\\ \textup{\textbf{Cosine Sum}}&\textbf{\textbf{Cosine Difference}}\\

\cos\left(a\right)+\cos\left(b\right)=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)

&\cos\left(a\right)-\cos\left(b\right)=-2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)\\\end{matrix}\]

Express the following sum or difference as a product containing only sines and/or cosines: \[\sin(7\theta)+\sin(3\theta)\]

Since it's sine being added by sine, this means that the you are going to be using \[\sin\left(a\right)+sin\left(b\right)=2sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)\] in order to express the sum as a product.So, with that being said, here is how to express the given sum: \[\begin{matrix} \sin\left(a\right)+\sin\left(b\right)&=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)\\\\ \sin\left(7\theta \right)+\sin\left(3\theta \right)&=2\sin\left(\frac{7\theta +3\theta }{2}\right)\cos\left(\frac{7\theta -3\theta }{2}\right)\\\\ &=2\sin\left(\frac{10\theta }{2}\right)\cos\left(\frac{4\theta }{2}\right)\\\\ &=\boxed{\mathbf{{\color{red} 2\sin\left(5\theta \right)+\cos\left(2\theta \right) }}} \end{matrix}\]

## Section L. - Find the Indicated Trigonometric Functions of the Given Angle

When it comes to finding the indicated trigonometric functions of a given angle, you are given 2 sides of a right triangle along with having to give the exact answers with rational denominators. This commonly means that the Pythagorean Theorem will be used.

Two sides of a right triangle ABC (C is the angle angle) are given. Find the indicated trigonometric function of the given angle. Give exact answers with rational denominators. \[\textup{Find} \sin A\: \textup{when}\: a=5\: \textup{and}\: b=8.\]

In order to find \(\sin\left(A\right)\), you need to recall that \(\sin\left(A \right)=\frac{ac}{a^2+b^2}\). However, since you are given \textup{sides a and b}, which means you need to find \textup{side c}, which means using the Pythagorean Theorem to find \textup{side c}, which means doing the following: \[c=\sqrt{a^2+b^2}=\sqrt{5^2+8^2}=\sqrt{25+64}=\sqrt{89}\]So, now that have all \(3\) sides, you can now find \(\sin\left(A\right)\), which means doing the following: \[\sin\left(A \right)=\frac{ac}{a^2+b^2}=\frac{5\sqrt{89}}{5^2+8^2}=\frac{5\sqrt{89}}{25+64}=\boxed{\mathbf{{\color{red} \frac{5\sqrt{89}}{89} }}}\]

## Section M. - Find the Exact Value of a Trigonometric Function

This is similar to Section D, but now you're being asked to find the exact value of the original trigonometric instead of using a double-angle formula while not dealing with an interval. However, as similar to Section L, you do need to use the Pythagorean Theorem in order to help you find the exact value of the indicated function.

Use the definition or identities to find the exact value of the following indicated trigonometric function of the acute angle \(\theta\): \[\begin{matrix} \sin(\theta)=\frac{3\sqrt{10}}{10}&&\textup{Find} \tan(\theta). \end{matrix}\]

So, recall that \(\sin\theta =\frac{3\sqrt{10}}{10}=\frac{o}{h}\) with \(\tan\theta =\frac{o}{a}\), which means that we are missing \(a\), which is the adjacent side. So, you can use the Pythagorean Theorem to find \(a\), which means that you can do the following: \[\begin{matrix} c=\sqrt{a^2+b^2}\\ (c)^2=(\sqrt{a^2+b^2})^2\\ c^2=a^2+b^2\\ c^2-b^2=a^2\\ a=\sqrt{c^2-b^2}\\ a=\sqrt{h^2-o^2}\\ a=\sqrt{10^2-(3\sqrt{10})^2}\\ a=\sqrt{100-9(10)}\\ a=\sqrt{100-90}\\ a=\sqrt{10} \end{matrix}\]So, now we know that \(\left\{\begin{matrix} a&=&\sqrt{10}\\o&=&3\sqrt{10}\\h&=&10 \end{matrix}\right.\). With that being said, you can now find \(\tan\theta\), which means doing the following: \[\tan\theta=\frac{o}{a}=\frac{3\sqrt{10}}{\sqrt{10}}=\frac{3(1)}{(1)}=\boxed{\mathbf{{\color{red} 3 }}}\]

## Section N. - Find the Exact Value of an Expression

When it comes to finding the exact value of an inverse expression, you are basically being told the angle whose trigonometric function is "\(x\)". For example, \(sin^{-1}\left(1\right)\) means "the angle whose sine is \(1\)". Afterwards, you are then given a trigonometric function for where you are then going to deal with trivial identities in order to solve it.

Find the exact value of the following expression: \[\sin^{-1}\frac{\sqrt{2}}{2}\]

\(\sin^{-1}\frac{\sqrt{2}}{2}\) means "the angle whose sine is \(\frac{\sqrt{2}}{2}\)", which means that \(\sin\theta=\frac{\sqrt{2}}{2}\), which means \(\sin\left ( 45^{\circ} \right )\), which is equivalent to \(\boxed{\mathbf{{\color{red} \frac{\pi}{4} }}}.\)

## Section O. - Naming the Quadrant

When it comes to naming the quadrant based on the two given trigonometric inequalities, you have to know the 4 quadrants and how trigonometry places a roll into which are positive and which are negative, which can all expressed with the following: \[\begin{matrix} \textup{\textbf{Quadrant I}}&\textup{Positive}&sin&cos&tan&csc&sec&cot\\ &\textup{Negative}&&&&&&\\\\

\textup{\textbf{Quadrant II}}&\textup{Positive}&sin&&&csc&&\\ &\textup{Negative}&&cos&tan&&sec&cot\\\\

\textup{\textbf{Quadrant III}}&\textup{Positive}&&&tan&&&cot\\ &\textup{Negative}&sin&cos&&csc&sec&\\\\

\textup{\textbf{Quadrant IV}}&\textup{Positive}&&cos&&&sec&\\ &\textup{Negative}&sin&&tan&csc&&cot \end{matrix}\]

Whatever the overlapping, or repeating, quadrant is, that will always be the right answer.

Name the quadrant in which angle \(\theta\) lies. \[\begin{matrix} \sin(\theta)>0&\textup{and}&\cos(\theta)<0 \end{matrix}\]

If \(\sin\theta >0\), then it's positive in Quadrants I and II.If \(\cos\theta <0\), then it's negative in Quadrants II and III.

Therefore, \(\sin\theta >0\) and \(\cos\theta <0\) both lie in \(\boxed{\mathbf{{\color{red} Quadrant\: II }}}\).

## Section P. - Solving a Triangle

When it comes to solving a triangle, you are given multiple choice answers with \(2\) given angles along with \(1\) side. One thing is for sure is that the law of sines is required to figure out the missing signs. To find the \(3^{rd}\) angle, just use the triangle sum theorem, which both formulas are the following: \[\begin{matrix} \textup{Law of Sines}&&\textup{Triangle Sum Theorem}\\\\ \frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}&&\measuredangle A+\measuredangle B+\measuredangle C=180^{\circ} \end{matrix}\]

However, I created a method that (currently) works when it comes figuring out \[\begin{matrix} \textup{What if:}\left\{\begin{matrix} \textup{side a} & = & \measuredangle B+\measuredangle C&&\textup{Highest Sum = Lowest Side}\\ \textup{side b} & = & \measuredangle A+\measuredangle C&&\textup{Middle Sum = Middle Side}\\ \textup{side c} & = & \measuredangle A+\measuredangle B&&\textup{Lowest Sum = Highest Side} \end{matrix}\right. \end{matrix}\]

However, if \(\textup{side a}=\textup{side b}=\textup{side c}\), then you can conclude that no triangle can exist.

Solve the following triangle:

First off, we need to find \(\measuredangle B\), which can be done by doing the following: \[\begin{matrix} \measuredangle a+\measuredangle B+\measuredangle C=180^{\circ}\\ 50^{\circ}+\measuredangle B+95^{\circ}=180^{\circ}\\ \measuredangle B+145^{\circ}=180^{\circ}\\ \boxed{\mathbf{{\color{red} \measuredangle B=35^{\circ} }}} \end{matrix}\]Since we have all \(3\) angles, we can now use the law of sines in order to find out \(\textup{sides a and c}\), which can be done by doing the following: \[\begin{matrix} \textup{side a}&\textup{side c}\\ \frac{a}{\sin A}=\frac{b}{\sin B}&\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ \frac{a}{\sin 50^{\circ}}=\frac{6}{\sin 35^{\circ}}&\frac{6}{\sin 35^{\circ}}=\frac{c}{\sin 95^{\circ}}\\\\ a\sin 35^{\circ}=6\sin 50^{\circ}&6\sin 95^{\circ}=c\sin 35^{\circ}\\ a=\frac{6\sin 50^{\circ}}{\sin 35^{\circ}}&\frac{6\sin 95^{\circ}}{\sin 35^{\circ}}=c\\ a=8.013346378&10.42087473=c\\ a=8.013346378&c=10.42087473\\ \boxed{\mathbf{{\color{red} a\approx 8.01 }}}&\boxed{\mathbf{{\color{red} c\approx 10.42 }}} \end{matrix}\]

However, we can use my created method to confirm that we did everything correctly, which can be noted by first doing the following \[\begin{matrix} \textup{What if:}\left\{\begin{matrix} \textup{side a} & = & \measuredangle B+\measuredangle C&=&35^{\circ}+95^{\circ}&=&130^{\circ}&&a=8.01\\ \textup{side b} & = & \measuredangle A+\measuredangle C&=&50^{\circ}+95^{\circ}&=&145^{\circ}&&b=6\\ \textup{side c} & = & \measuredangle A+\measuredangle B&=&50^{\circ}+35^{\circ}&=&85^{\circ}&&c=10.42 \end{matrix}\right. \end{matrix}\]

Now, we can then confirm the sides about doing the following: \[\begin{matrix} 85^{\circ}&<&130^{\circ}&<&145^{\circ}\\ c&<&a&<&b\\ 10.42&>&8.01&>&6\\ \end{matrix}\]

This is now both true and correct.

## Section Q. - Solving a (Possible) Triangle

When it comes to solving for a possible triangle, this is similar to Sections G and P, but you're given 2 angles and a side. based on the info that you are given, it can be for either one triangle or no triangle. However, it's use my created method from Section P to look more into this. But first, let's use the Pythagorean Theorem to figure out the remaining side.

Two sides and an angle are given. Determine whether the following given information results in one triangle, two triangles, or no triangle. Solve any triangle(s) that results: \[\begin{matrix} b=4&c=8&B=65^{\circ} \end{matrix}\]

\[\begin{matrix} \textup{A})\: \textup{One Triangle}:\left\{\begin{matrix} C=32^{\circ}\\A=83^{\circ}\\a=14 \end{matrix}\right. && \textup{B})\: \textup{One Triangle}:\left\{\begin{matrix} C=34^{\circ}\\A=81^{\circ}\\a=16 \end{matrix}\right. &&\\ \\ \textup{C})\: \textup{One Triangle}:\left\{\begin{matrix} C=33^{\circ}\\A=82^{\circ}\\a=12 \end{matrix}\right. && \textup{D})\: \textup{No Triangle} \end{matrix}\]

First off, we need to find \(\textup{side a}\), which can be done by doing the following: \[\begin{matrix} c=\sqrt{a^2+b^2}\\ (c)^2=(\sqrt{a^2+b^2})^2\\ c^2=a^2+b^2\\ c^2-b^2=a^2\\ a=\sqrt{c^2-b^2}\\ a=\sqrt{h^2-o^2}\\ a=\sqrt{8^2-4^2}\\ a=\sqrt{64-16}\\ a=\sqrt{48}\\ a=\sqrt{2^4\cdot 3}\\ a=4\sqrt{3}\\ a\approx 6.93 \end{matrix}\]Now, based on the given multiple choices, none of them match what \(\textup{side a}\) correctly is. Therefore, we can conclude that none of the given choices are correctly, which means that \(\boxed{\mathbf{{\color{red} no\: triangle }}}\) can be made.

## Section R. - Finding the Exact Value of the Indicated Trigonometric Function

This is similar to Sections D, M, and O. It's similar to Section D in terms of finding the value while dealing with an interval of \(\theta\). It's similar to Section M in terms of using a definition or identity in order to find the exact value along with the fact that the Pythagorean Theorem will be used thanks to Soc-Cah-Toa or even an identity. It's similar to Section O in terms of finding where \(\theta\) lies.

Find the exact value of the following indicated trigonometric function: \[\begin{matrix} \cos(\theta)=\frac{4}{7},&\tan(\theta)<0&&&\textup{Find} \sin(\theta). \end{matrix}\]

\[\boxed{\textup{P.S. You would have to use the Pythagorean Theorem in order to solve this problem.}}\]

If \(\tan\theta <0\), then it's negative in Quadrants II and IV, which means that this will have a negative in the front. Now, you can recall the following: \[\begin{matrix} \cos\theta &=&\frac{4}{7}&=\frac{a}{h}\\ \sin\theta &&&=\frac{o}{h}\\ \end{matrix}\] and this helps indicate that we are missing \textup{side b}, which is the \textup{opposite side}, which can be done by doing the following: \[\begin{matrix} c=\sqrt{a^2+b^2}\\ (c)^2=(\sqrt{a^2+b^2})^2\\ c^2=a^2+b^2\\ c^2-a^2=b^2\\ b=\sqrt{c^2-a^2}\\ o=\sqrt{h^2-a^2}\\ o=\sqrt{7^2-4^2}\\ o=\sqrt{49-16}\\ o=\sqrt{33} \end{matrix}\]Thanks to \(\tan\theta <0\) going to have a negative effect, if \(\sin\theta=\frac{o}{h}\), then \(-\sin\theta=-\frac{o}{h}\), which now means you just substitute and doing the following to find the exact value: \[-\sin\theta=-\frac{o}{h}=\boxed{\mathbf{{\color{red} -\frac{\sqrt{33}}{7} }}}\]

Find the exact value of the following indicated trigonometric function: \[\begin{matrix} \cot(\theta)=-\frac{7}{2},&\cos(\theta)<0&&&\textup{Find} \csc(\theta). \end{matrix}\]

\[\boxed{\textup{P.S. You would have to use identities in order to solve this problem.}}\]

Well, for this problem, there were 2 different ways to solve this problem while using the same identity with that identity being \(1+\cot^2\theta=\csc^2\theta\). One method can be using substituting early or substituting towards to the end.\[\begin{matrix} \textup{\textbf{Method 1}}&&\textup{\textbf{Method 2}}\\ \begin{matrix} 1+\cot^2\theta=\csc^2\theta\\ 1+\left(\frac{7}{2}\right)^2=\csc^2\theta\\ 1+\frac{49}{4}=\csc^2\theta\\ \frac{4}{4}+\frac{49}{4}=\csc^2\theta\\ \frac{53}{4}=\csc^2\theta\\ \frac{53}{4}=\frac{1}{\sin^2\theta}\\ \frac{53}{4}\sin^2\theta=1\\ \sin^2\theta=\frac{4}{53}\\ \sqrt{\sin^2\theta}=\sqrt{\frac{4}{53}}\\ \sin\theta=\pm \sqrt{\frac{4}{53}}\\ \sin\theta=\sqrt{\frac{4}{53}}\\ \sin\theta=\frac{2}{\sqrt{53}}\\ \frac{1}{\csc\theta}=\frac{2}{\sqrt{53}}\\ 1=\frac{2}{\sqrt{53}}\csc\theta\\ \frac{\sqrt{53}}{2}=\csc\theta\\ \boxed{\mathbf{{\color{red} \csc\theta=\frac{\sqrt{53}}{2} }}} \end{matrix}&&\begin{matrix} 1+\cot^2\theta=\csc^2\theta\\ 1+\cot^2\theta=\frac{1}{\sin^2\theta}\\ \sin^2\theta(1+\cot^2\theta)=1\\ \sin^2\theta=\frac{1}{1+\cot^2\theta}\\ \sqrt{\sin^2\theta}=\sqrt{\frac{1}{1+\cot^2\theta}}\\ \sin\theta=\pm\sqrt{\frac{1}{1+\cot^2\theta}}\\ \sin\theta=\sqrt{\frac{1}{1+\cot^2\theta}}\\ \sin\theta=\sqrt{\frac{1}{1+\left ( -\frac{7}{2} \right )^2}}\\ \sin\theta=\sqrt{\frac{1}{1+\frac{49}{4} }}\\ \sin\theta=\sqrt{\frac{1}{\frac{4}{4}+\frac{49}{4} }}\\ \sin\theta=\sqrt{\frac{1}{\frac{53}{4} }}\\ \sin\theta=\sqrt{\frac{4}{53}}\\ \sin\theta=\frac{2}{\sqrt{53}}\\ \frac{1}{\csc\theta}=\frac{2}{\sqrt{53}}\\ 1=\frac{2}{\sqrt{53}}\csc\theta\\ \frac{\sqrt{53}}{2}=\csc\theta\\ \boxed{\mathbf{{\color{red} \csc\theta=\frac{\sqrt{53}}{2} }}} \end{matrix} \end{matrix}\]

## Section S. - Find the Exact Value of an Expression - Part 2

When it comes to finding the exact value of an expression, trivial identities are sometimes required in order to solve some of them. However, for other expressions, you can treat the trigonometric art of the expression as if it was a base and exponent like using the following exponential rule: \[a^b\cdot a^c=a^{b+c}.\]

Other times, it's basic wording to find out the exact value of the given expression, but it's based on the expression.

Find the exact value of the following expression. Do not use a calculator. \[1+\cot^2 30^{\circ}-\sec^2 45^{\circ}\] \[\boxed{\textup{P.S. This example is for when you are given an expression that's in degrees.}}\]

\[\begin{matrix} 1+\cot ^2\left(30^{\circ }\right)-\sec ^2\left(45^{\circ }\right)\\ =1+\cot ^2\left(30^{\circ }\right)-\frac{1}{\cos ^2\left(45^{\circ }\right)}\\ =1+\left(\sqrt{3}\right)^2-\frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2}\\ =1+3-\frac{1}{\left(\frac{2}{4}\right)}\\ =1+3-\frac{1}{\frac{1}{2}}\\ =1+3-2\\ =4-2\\ =\boxed{\mathbf{{\color{red} 2 }}} \end{matrix}\]

Find the exact value of the following expression. Do not use a calculator. \[\sin \frac{\pi }{3}-\cos\frac{\pi }{6}\] \[\boxed{\textup{P.S. This example is for when you are given an expression that's in radians.}}\]

There are two ways in order to solve this trigonometric expression. One method can be using an identity with another identity can be using trivial identities.\(\textup{\textbf{Method 1: Using an Identity}}\)

If you want, you can note that \(\sin \left(x\right)=\cos \left(\frac{\pi }{2}-x\right)\), which means that \(\sin \left(\frac{\pi }{3}\right)=\cos \left(\frac{\pi }{2}-\frac{\pi }{3}\right)\), which means that you can simplify this as the following: \[\begin{matrix} \sin \left(\frac{\pi }{3}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\cos \left(\frac{\pi }{2}-\frac{\pi }{3}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\cos \left(\frac{\pi (3)}{2(3)}-\frac{\pi (2)}{3(2)}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\cos \left(\frac{3\pi }{6}-\frac{2\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\cos \left(\frac{(3-2)\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\cos \left(\frac{\pi }{6}\right)-\cos \left(\frac{\pi }{6}\right)\\ =(1-1)\cos \left(\frac{\pi }{6}\right)\\ =(0)\cos \left(\frac{\pi }{6}\right)\\ =\boxed{\mathbf{{\color{red} 0 }}} \end{matrix}\]

\(\textup{\textbf{Method 2: Using Trivial Identities}}\) \[\begin{matrix} \sin \left(\frac{\pi }{3}\right)-\cos \left(\frac{\pi }{6}\right)\\ =\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\\\\ =\frac{\sqrt{3}-\sqrt{3}}{2}\\\\ =\frac{(1-1)\sqrt{3}}{2}\\\\ =\frac{(0)\sqrt{3}}{2}\\\\ =\frac{0}{2}\\\\ =\boxed{\mathbf{{\color{red} 0 }}} \end{matrix}\]

Find the exact value of the following expression. Do not use a calculator. \[\sin[\sin^{-1}(-0.2)]\] \[\boxed{\textup{P.S. I will be using an exponential rule in order to solve this problem.}}\]

\[\begin{matrix} \sin[\sin^{-1}(-0.2)]\\ =\sin^1[\sin^{-1}(-0.2)]\\ =\sin^{1-1}(-0.2)\\ =\sin^{0}(-0.2)\\ =(1)(-0.2)\\ =\boxed{\mathbf{{\color{red} -0.2 }}} \end{matrix}\]The funny thing about this expression is that \[\sin[\sin^{-1}(x)]=\sin^{-1}[\sin(x)]=x\] which can be a fun little trick with regards to solving this trigonometric expression.

Find the exact value of the following expression. Do not use a calculator. \[\cos[\cos^{-1}(-0.9372)]\] \[\boxed{\textup{P.S. I will be using words in order to solve this problem.}}\]

\(\cos[\cos^{-1}(-0.9372)]\) means that the cosine of an angle whose cosine is \(-0.9372\). So, if the angle that has a cosine of \(-0.9372\) is \(x\), then \(\cos(x)=\boxed{\mathbf{{\color{red} -0.9372 }}}\).The funny thing about this expression is that \[\cos[\cos^{-1}(x)]=\cos^{-1}[\cos(x)]=x\] which can be a fun little trick with regards to solving this trigonometric expression.

## Section T. - Find the Exact Value of an Expression - Part 3

This is the same concept as Section S, but it's slightly different when it comes to dealing with radians and degrees in trigonometry. When it comes to radians, you can simplify it by being aware that \(\pi=180^{\circ}\) then using an identity. When it comes to degrees, there are various ways into solving them. Just make sure it is fully simplified if not factored. But either way, there are various methods with regards to finding the exact value of the trigonometric expression.

Find the exact value of the following expression: \[\sin \frac{\pi}{12}\] \[\boxed{\textup{P.S. This example is for when you are given an expression that's in radians.}}\]

\(\textup{\textbf{Method 1: Using a Summation Identity}}\)\[\begin{matrix} \sin \left(\frac{\pi }{12}\right)&=\sin \left(\frac{180^{\circ} }{12}\right)&=\sin \left(15^{\circ}\right)\\ &&=\sin \left(45^{\circ}-30^{\circ}\right)\\ &&=\sin \left(45^{\circ}\right)\cos \left(30^{\circ}\right)-\cos \left(45^{\circ}\right)\sin \left(30^{\circ}\right)\\ &&=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot \frac{1}{2}\\ &&=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}\\ &&=\frac{\sqrt{3}-1}{2\sqrt{2}}\\ &&=\frac{\sqrt{3}-1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\\ &&=\frac{\sqrt{2}(\sqrt{3}-1)}{2\sqrt{4}}\\ &&=\frac{\sqrt{2}(\sqrt{3}-1)}{2(2)}\\ &&=\boxed{\mathbf{{\color{red} \frac{\sqrt{2}(\sqrt{3}-1)}{4} }}} \end{matrix}\]

\(\textup{\textbf{Method 2: Using a Half-Angle Identity}}\) \[\begin{matrix} \sin \left(\frac{\pi }{12}\right)&=\sin \left(\frac{\frac{\pi }{6}}{2}\right)\\ &=\sqrt{\frac{1-\cos \left(\frac{\pi }{6}\right)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{\frac{2}{2}-\frac{\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt{3}}{4}}\\ &=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}} &=\frac{\sqrt{2-\sqrt{3}}}{2} \end{matrix}\]

The boxed one in red in \textup{Method 1} is the answer that is listed in the actual study guide while the answer in \textup{Method 2} is an alternative answer with regards to solving this problem.

Find the exact value of the following expression: \[\tan 345^{\circ}\] \[\boxed{\textup{P.S. This example is for when you are given an expression that's in degrees.}}\]

\(\textup{\textbf{Method 1: Simplify the Express First}}\) \[\begin{matrix} \tan \left(345^{\circ }\right)&=\tan \left(345^{\circ }-180^{\circ }\right)\\ &=\tan \left(165^{\circ }\right)\\ &=\tan \left(135^{\circ }+30^{\circ }\right)\\ &=\frac{\tan\left(135^{\circ }\right)+\tan\left(30^{\circ }\right)}{1-\tan\left(135^{\circ }\right)\tan\left(30^{\circ }\right)}\\ &=\frac{-1+\frac{\sqrt{3}}{3}}{1-\left(-1\right)\left(\frac{\sqrt{3}}{3}\right)}\\ &=\frac{-1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}\\ &=\frac{\frac{-\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\\ &=\frac{\left(-\sqrt{3}+1\right)\sqrt{3}}{\sqrt{3}\left(\sqrt{3}+1\right)}\\ &=\frac{-\sqrt{3}+1}{\sqrt{3}+1}\\ &=\frac{\left(-\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ &=\frac{2\sqrt{3}-4}{2}\\ &=\frac{2\left(\sqrt{3}-2\right)}{2}\\ &=\sqrt{3}-2\\ &=\boxed{\mathbf{{\color{red} -2+\sqrt{3} }}} \end{matrix}\]\(\textup{\textbf{Method 2: Using a Summation Identity First}}\) \[\begin{matrix} \tan \left(345^{\circ }\right)&=\tan \left(300^{\circ }-45^{\circ }\right)\\ &=\frac{\tan\left(300^{\circ }\right)+\tan\left(45^{\circ }\right)}{1-\tan\left(300^{\circ }\right)\tan\left(45^{\circ }\right)}\\ &=\frac{-\sqrt{3}+1}{1-\left(\sqrt{3}\right)\left(1\right)}\\ &=\frac{1-\sqrt{3}}{1+\sqrt{3}}\\ &=\frac{1-\sqrt{3}}{1+\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)\\ &=\frac{1-\sqrt{3}-\sqrt{3}+3}{1-\sqrt{9}}\\ &=\frac{4-2\sqrt{3}}{1-3}\\ &=\frac{4-2\sqrt{3}}{(-2)}\\ &=-\frac{2\left(2-\sqrt{3}\right)}{2}\\ &=-(2-\sqrt{3})\\ &=\boxed{\mathbf{{\color{red} -2+\sqrt{3} }}} \end{matrix}\]

## Section U. - Using Definitions and Identities

This is literally the exact same concept as Section M. You may need to use the Pythagorean Theorem to solve them while others may just require an identity to simplify everything that was given in order to find the exact value of the indicated trigonometric function. In some cases, there are more than one way to solve various problems.

Use the definition or identities to find the exact value of the following indicated trigonometric function of the acute angle \(\theta\): \[\begin{matrix} \cos(\theta)=\frac{\sqrt{10}}{10}&&\textup{Find} \sec(\theta). \end{matrix}\] \[\boxed{\textup{P.S. You would have to use a definition in order to solve this problem.}}\]

If you can recall that \(\sec \theta =\frac{1}{\cos \theta }\), which means you have to flip the given indicated cosine function then simplify, which means doing the following: \[\sec \theta =\frac{1}{\cos\theta }=\frac{1}{\frac{\sqrt{10}}{10}}=\frac{10}{\sqrt{10}}=\frac{10}{\sqrt{10}}\left(\frac{\sqrt{10}}{\sqrt{10}}\right)=\frac{10\sqrt{10}}{\sqrt{100}}=\frac{10\sqrt{10}}{10}=\boxed{\mathbf{{\color{red} \sqrt{10} }}}\]

Use the definition or identities to find the exact value of the following indicated trigonometric function of the acute angle \(\theta\): \[\begin{matrix} \cos(\theta)=\frac{5}{13}&&\textup{Find} \cot(\theta). \end{matrix}\] \[\boxed{\textup{P.S. You would have to use an identity in order to solve this problem.}}\]

There are ironically two different methods with regards to solve this problem. One method involves using the Pythagorean Theorem while another method involves using an if-then scenario then substitution.\(\textup{\textbf{Method 1: Pythagorean Theorem}}\)

Now, you can recall the following: \[\begin{matrix} \cos\theta &=&\frac{5}{13}&=\frac{a}{h}\\ \sin\theta &&&=\frac{o}{h}\\ \end{matrix}\] and this helps indicate that we are missing \textup{side b}, which is the \textup{opposite side}, which can be done by doing the following: \[\begin{matrix} c=\sqrt{a^2+b^2}\\ (c)^2=(\sqrt{a^2+b^2})^2\\ c^2=a^2+b^2\\ c^2-a^2=b^2\\ b=\sqrt{c^2-a^2}\\ o=\sqrt{h^2-a^2}\\ o=\sqrt{13^2-5^2}\\ o=\sqrt{169-25}\\ o=\sqrt{144}\\ o=12 \end{matrix}\]

So, now we know that \(\left\{\begin{matrix} a&=&5\\o&=&12\\h&=&13 \end{matrix}\right.\). With that being said, we can now find \(\cot\theta\), which means doing the following: \[\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{o}{a}}=\frac{a}{o}=\boxed{\mathbf{{\color{red} \frac{5}{12} }}}\]

\(\textup{\textbf{Method 2: If-Then Scenario}}\)

For this scenario, if \(\cot \theta =\pm \frac{1}{\sqrt{\sec^2\theta -1}}\) and \( sec\theta=\frac{1}{\cos\theta}\), then this would mean the following: \[\cot\theta =\pm \frac{1}{\sqrt{\frac{1}{\cos^2\theta }-1}}\]

So, since \(\cot\theta\) is positive, this means that you will only be using the positive side of \(\frac{1}{\sqrt{\frac{1}{\cos^2\theta }-1}}\) in order to solve this problem. Now, since \(\cos\theta=\frac{5}{13}\), then this would mean the following: \[\begin{matrix} \frac{1}{\sqrt{\frac{1}{\cos^2\theta }-1}}=\frac{1}{\sqrt{\frac{1}{\left ( \frac{5}{13} \right )^2 }-1}}=\frac{1}{\sqrt{\frac{1}{ \frac{25}{169} }-1}}=\frac{1}{\sqrt{\frac{169}{25}-1}}\\\\ =\frac{1}{\sqrt{\frac{169}{25}-\frac{25}{25}}}=\frac{1}{\sqrt{\frac{144}{25}}}=\frac{1}{\frac{12}{5}}=\boxed{\mathbf{{\color{red} \frac{5}{12} }}} \end{matrix}\]

## Section V. - Solving Equations on an Interval

When it comes to problems dealing with solving equations an an interval, the interval is always going to be \(\textup{angle}\:\theta=0\le \theta \le 2\pi\) 9 out of 10 times especially since \(2\pi=360^{\circ}\). In most terms, it's best to use substituting in order to solve the equations. Just remember to replug the original substitution in order to properly solve the equations.

Commonly, you should "\(x\)" as your substitution value. In other cases, some use "\(u\)" for substitution since "\(u\)-substitution" is a term taught in some classes. Either way, just be ready to substitute and even substitute back once everything is simplified in its substitute form.

Solve the following equation on the interval \(0\le \theta \le 2\pi\): \[\cos^2\theta+2\cos\theta+1=0\]

\(\textup{\textbf{Method 1: U-Substitution}}\)When it comes to \(u\)-substitution, you first let \(u=\cos\theta\), which means that \(\left\{\begin{matrix} \textup{If}\: u=\cos\theta\\ \textup{Then}\: u^2=\cos^2\theta\\ \end{matrix}\right. .\) This means the following would start off the first half of this problem: \[\begin{matrix} \cos^2\theta+2\cos\theta+1=0\\ u^2+2u+1=0\\ u^2+u+u+1=0\\ (u^2+u)+(u+1)=0\\ u(u+1)+1(u+1)=0\\ (u+1)(u+1)=0\\ (u+1)^2=0\\ \sqrt{(u+1)^2}=\sqrt{0}\\ u+1=0\\ u=-1\\ \end{matrix}\]

Now that the \(u\)-substitution is done, you have to plug \(\cos\theta\) back in since we let \(u=\cos\theta\) and now find \(\theta\), which means doing the following: \[\begin{matrix} u=-1\\\\ \cos\theta=-1\\ \cos\left ( \pi \right )=-1\\\\ \theta=\pi\\\\ \boxed{\mathbf{{\color{red} \left \{ \pi \right \} }}} \end{matrix}\]

\(\textup{\textbf{Method 2: Regular Simplifying }}\)

\[\begin{matrix} \cos^2\theta+2\cos\theta+1=0\\ \cos^2+\cos\theta+\cos\theta+1=0\\ (\cos^2\theta+\cos\theta)+(\cos\theta+1)=0\\ \cos\theta(\cos\theta+1)+1(\cos\theta+1)=0\\ (\cos\theta+1)(\cos\theta+1)=0\\ (\cos\theta+1)^2=0\\ \sqrt{(\cos\theta+1)^2}=\sqrt{0}\\ \cos\theta+1=0\\ \cos\theta=-1\\ \cos\left ( \pi \right )=-1\\\\ \theta=\pi\\\\ \boxed{\mathbf{{\color{red} \left \{ \pi \right \} }}} \end{matrix}\]

Solve the following equation on the interval \(0\le \theta \le 2\pi\): \[2\sin^2\theta=\sin\theta\]

\(\textup{\textbf{Method 1: U-Substitution}}\)When it comes to \(u\)-substitution, you first let \(u=\sin\theta\), which means that \(\left\{\begin{matrix} \textup{If}\: u=\sin\theta\\ \textup{Then}\: u^2=\sin^2\theta\\ \end{matrix}\right. .\) This means the following would start off the first half of this problem: \[\begin{matrix} &2\sin ^2\theta =\sin \theta& \\ &2u^2 =u&\\ &2u^2-u=u-u&\\ &2u^2-u=0&\\ &u(2u-1)=0&\\ u=0&&2u-1=0\\ &&2u-1+1=0+1\\ &&2u=1\\ &&\frac{2u}{2}=\frac{1}{2}\\ &&u=\frac{1}{2} \end{matrix}\]

Now that the \(u\)-substitution is done, you have to plug \(\sin\theta\) back in since we let \(u=\sin\theta\) and now find \(\theta\), which means doing the following: \[\begin{matrix} u=0&&u=\frac{1}{2}\\\\ \sin\theta=0&&\sin\theta=\frac{1}{2}\\\\ \sin\left ( 0 \right )=0&&\sin\left ( \frac{\pi}{6} \right )=\frac{1}{2}\\ \sin\left ( \pi \right )=0&&\sin\left ( \frac{5\pi}{6} \right )=\frac{1}{2}\\\\ \left \{ 0,\pi \right \}&&\left \{ \frac{\pi}{6},\frac{5\pi}{6} \right \}\\\\ &\boxed{\mathbf{{\color{red} \left \{ 0,\pi,\frac{\pi}{6},\frac{5\pi}{6} \right \} }}}&

\end{matrix}\]

\(\textup{\textbf{Method 2: Regular Simplifying }}\)

\[\begin{matrix} &2\sin ^2\theta =\sin \theta& \\ &2\sin^2\theta-\sin\theta=\sin\theta-\sin\theta&\\ &2\sin^2\theta-\sin\theta=0&\\ &\sin\theta(2\sin\theta-1)=0&\\ \sin\theta=0&&2\sin\theta-1=0\\ &&2\sin\theta-1+1=0+1\\ &&2\sin\theta=1\\ &&\frac{2\sin\theta}{2}=\frac{1}{2}\\ &&\sin\theta=\frac{1}{2}\\\\ &&\sin\theta=\frac{1}{2}\\\\ \sin\left ( 0 \right )=0&&\sin\left ( \frac{\pi}{6} \right )=\frac{1}{2}\\ \sin\left ( \pi \right )=0&&\sin\left ( \frac{5\pi}{6} \right )=\frac{1}{2}\\\\ \left \{ 0,\pi \right \}&&\left \{ \frac{\pi}{6},\frac{5\pi}{6} \right \}\\\\ &\boxed{\mathbf{{\color{red} \left \{ 0,\pi,\frac{\pi}{6},\frac{5\pi}{6} \right \} }}}&\end{matrix}\]

## Section W. - Establish the Identity

When it comes to establishing an identity, you are basically being asked to "prove that the left side equals the right side". You establish the identity by using identities. The creative thing about establishing identities is that there are many different ways to solve them and prove that they are true. In some cases, you can simplify, also known as "manipulating", the left to prove the right. In other cases, you can manipulate the right to prove the left. However, you sometimes have to manipulate both sides at various times. To save time since I found several different ways (yes, I had a lot of time on my hand in order to find that way different ways for each given identity) to solve each of the given identities in this study guide, I will show the ones that I personally found. This is the main thing when it comes to establishing identities: \(\boxed{\textup{steps may vary}}\) thanks to the fact that there are many ways to establishing them.

Establish the following identity: \[\sec u+\tan u=\frac{\cos u}{1-\sin u}\]

\(\textup{\textbf{Method 1}}\)\[\begin{matrix} \sec u+\tan u=\frac{\cos u}{1-\sin u}\\\\ \frac{1}{\cos u}+\frac{\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ \frac{1+\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ (1+\sin u)(1-\sin u)=(\cos u)(\cos u)\\ 1-\sin^2 u=\cos^2 u\\ \cos^2 u=\cos^2 u\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

\(\textup{\textbf{Method 2}}\)

\[\begin{matrix} \sec u+\tan u=\frac{\cos u}{1-\sin u}\\\\ \frac{1}{\cos u}+\frac{\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ \frac{1+\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ \frac{1+\sin u}{\cos u}\cdot \frac{1-\sin u}{1-\sin u}=\frac{\cos u}{1-\sin u}\\\\ \frac{(1+\sin u)(1-\sin u)}{\cos u(1-\sin u)}=\frac{\cos u}{1-\sin u}\\\\ \frac{1-\sin^2 u}{\cos u(1-\sin u)}=\frac{\cos u}{1-\sin u}\\\\ \frac{\cos^2 u}{\cos u(1-\sin u)}=\frac{\cos u}{1-\sin u}\\\\ \frac{(\cos u)(\cos u)}{\cos u(1-\sin u)}=\frac{\cos u}{1-\sin u}\\\\ \frac{\cos u}{1-\sin u}=\frac{\cos u}{1-\sin u}\\\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

\(\textup{\textbf{Method 3}}\)

\[\begin{matrix} \sec u+\tan u=\frac{\cos u}{1-\sin u}\\\\ \frac{1}{\cos u}+\frac{\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ \frac{1+\sin u}{\cos u}=\frac{\cos u}{1-\sin u}\\\\ \frac{1+\sin u}{\cos u}=\frac{\cos u}{\frac{1-\sin^2 u}{1+\sin u}}\\\\ \frac{1+\sin u}{\cos u}=\frac{(\cos u)(1+\sin u)}{1-\sin^2 u}\\\\ \frac{1+\sin u}{\cos u}=\frac{(\cos u)(1+\sin u)}{\cos^2 u}\\\\ \frac{1+\sin u}{\cos u}=\frac{(\cos u)(1+\sin u)}{(\cos u)(\cos u)}\\\\ \frac{1+\sin u}{\cos u}=\frac{1+\sin u}{\cos u}\\\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

Establish the following identity: \[1-\frac{\cos^2 u}{1-\sin u}=-\sin u\]

\(\textup{\textbf{Method 1}}\)\[\begin{matrix} 1-\frac{\cos^2 u}{1-\sin u}=-\sin u\\\\ 1-\frac{1-\sin^2u}{1-\sin u}=-\sin u\\\\ 1-\frac{(1+\sin u)(1-\sin u)}{1-\sin u}=-\sin u\\\\ 1-(1+\sin u)=-\sin u\\ 1-1-\sin u=-\sin u\\ (1-1)-\sin u=-\sin u\\ (0)-\sin u=-\sin u\\ -\sin u=-\sin u\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

\(\textup{\textbf{Method 2}}\)

\[\begin{matrix} 1-\frac{\cos^2 u}{1-\sin u}=-\sin u\\\\ -\frac{\cos^2 u}{1-\sin u}=-1-\sin u\\\\ -\frac{\cos^2 u}{1-\sin u}=-(1+\sin u)\\\\ -\frac{\cos^2 u}{1-\sin u}=-\left ( \frac{1-\sin^2 u}{1-\sin u} \right )\\\\ -\frac{\cos^2 u}{1-\sin u}=-\left ( \frac{\cos^2 u}{1-\sin u} \right )\\\\ -\frac{\cos^2 u}{1-\sin u}=-\frac{\cos^2 u}{1-\sin u}\\\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

Establish the following identity: \[\frac{\cos u}{\cos u-\sin u}=\frac{1}{1-\tan u}\]

\(\textup{\textbf{Method 1}}\)\[\begin{matrix} \frac{\cos u}{\cos u-\sin u}=\frac{1}{1-\tan u}\\\\ \frac{\cos u}{\cos u-\sin u}\cdot \frac{\frac{1}{\cos u}}{\frac{1}{\cos u}}=\frac{1}{1-\tan u}\\\\ \frac{1}{1-\frac{\sin u}{\cos u}}=\frac{1}{1-\tan u}\\\\ \frac{1}{1-\tan u}=\frac{1}{1-\tan u}\\\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

\(\textup{\textbf{Method 2}}\)

\[\begin{matrix} \frac{\cos u}{\cos u-\sin u}=\frac{1}{1-\tan u}\\\\ \frac{\cos u}{\cos u-\sin u}=\frac{1}{1-\frac{\sin u}{\cos u}}\\\\ \frac{\cos u}{\cos u-\sin u}=\frac{1}{\frac{\cos u}{\cos u}-\frac{\sin u}{\cos u}}\\\\ \frac{\cos u}{\cos u-\sin u}=\frac{1}{\frac{\cos u-\sin u}{\cos u}}\\\\ \frac{\cos u}{\cos u-\sin u}=\frac{\cos u}{\cos u-\sin u}\\\\ \boxed{\mathbf{{\color{red} True }}} \end{matrix}\]

## Section X. - Simplifying the Trigonometric Expressions

This is somewhat the same concept as Section V in terms of using substituting \(x\), or \(u\) if using "\(u\)-substitution", then substituting back in order to simplify the expressions. Some will ask you to rewrite the expression as a common denominator while others may just ask you to factor and simplify.

Simplify the following trigonometric expression by the following direction: \[\textup{Rewrite over a common denominator:}\: \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\]

\(\textup{\textbf{Method 1: X-Substitution}}\)When it comes to \(x\)-substitution, you first let \(x=\sin\theta\), which means the following would start off the first half of this problem: \[\begin{matrix} \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}&=\frac{1}{1-x}+\frac{1}{1+x}\\\\ &=\frac{1}{1-x}\cdot \frac{1+x}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{1-x}\\\\ &=\frac{1+x}{1-x^2}+\frac{1-x}{1-x^2}\\\\ &=\frac{1+x+1-x}{1-x^2}\\\\ &=\frac{x-x+1+1}{1-x^2}\\\\ &=\frac{2}{1-x^2} \end{matrix}\]

Now that the \(x\)-substitution is done, you have to plug \(\sin\theta\) back in since we let \(x=\sin\theta\) and now find \(\theta\), which means doing the following: \[\frac{2}{1-x^2}=\frac{2}{1-\sin^2 \theta}=\boxed{\mathbf{{\color{red} \frac{2}{\cos^2 \theta} }}}\]

\(\textup{\textbf{Method 2: Regular Simplifying }}\)

\[\begin{matrix} \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}&=\frac{1}{1-\sin \theta}\cdot \frac{1+\sin \theta}{1+\sin \theta}+\frac{1}{1+\sin \theta}\cdot \frac{1-\sin \theta}{1-\sin \theta}\\\\ &=\frac{1+\sin \theta}{1-\sin \theta}+\frac{1-\sin \theta}{1-\sin \theta}\\\\ &=\frac{1+\sin \theta+1-\sin \theta}{1-\sin^2 \theta}\\\\ &=\frac{\sin \theta-\sin \theta+1+1}{1-\sin^2 \theta}\\\\ &=\frac{2}{1-\sin^2 \theta}\\\\ &=\boxed{\mathbf{{\color{red} \frac{2}{\cos^2 \theta} }}} \end{matrix}\]

Simplify the following trigonometric expression by the following direction: \[\textup{Factor and simplify:}\: \frac{5\cos^2 \theta+6\cos \theta+1}{\cos^2 \theta -1}\]

\(\textup{\textbf{Method 1: X-Substitution}}\)When it comes to \(x\)-substitution, you first let \(x=\cos\theta\), which means that \(\left\{\begin{matrix} \textup{If}\: x=\cos\theta\\ \textup{Then}\: x^2=\cos^2\theta\\ \end{matrix}\right. .\) This means the following would start off the first half of this problem: \[\begin{matrix} \frac{5\cos^2 \theta+6\cos \theta+1}{\cos^2 \theta -1}&=\frac{5x^2+6x+1}{x^2-1}\\\\ &=\frac{5x^2+5x+x+1}{x^2-1^2}\\\\ &=\frac{(5x^2+5x)+(x+1)}{\left(x+1\right)\left(x-1\right)}\\\\ &=\frac{5x(x+1)+1(x+1)}{\left(x+1\right)\left(x-1\right)}\\\\ &=\frac{\left(5x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\\\ &=\frac{\left(5x+1\right)\left(1\right)}{\left(1\right)\left(x-1\right)}\\\\ &=\frac{5x+1}{x-1} \end{matrix}\]

Now that the \(x\)-substitution is done, you have to plug \(\cos\theta\) back in since we let \(x=\cos\theta\) and now find \(\theta\), which means doing the following: \[\frac{5x+1}{x-1}=\boxed{\mathbf{{\color{red} \frac{5\cos \theta+1}{\cos \theta-1} }}}\]

\(\textup{\textbf{Method 2: Regular Simplifying }}\)

\[\begin{matrix} \frac{5\cos^2 \theta+6\cos \theta+1}{\cos^2 \theta -1}&=\frac{5\cos^2 \theta+5\cos \theta+\cos \theta+1}{\cos^2 \theta-1^2}\\\\ &=\frac{(5\cos^2 \theta+5\cos \theta)+(\cos \theta+1)}{\left(\cos \theta+1\right)\left(\cos \theta-1\right)}\\\\ &=\frac{5\cos \theta(\cos \theta+1)+1(\cos \theta+1)}{\left(\cos \theta+1\right)\left(\cos \theta-1\right)}\\\\ &=\frac{\left(5\cos \theta+1\right)\left(\cos \theta+1\right)}{\left(\cos \theta+1\right)\left(\cos \theta-1\right)}\\\\ &=\frac{\left(5\cos \theta+1\right)\left(1\right)}{\left(1\right)\left(\cos \theta-1\right)}\\\\ &=\boxed{\mathbf{{\color{red} \frac{5\cos \theta+1}{\cos \theta-1} }}} \end{matrix}\]

## Section Y. - Solving the Equation on an Interval

This is the exact same concept as Section V. Just simplify then use trivial identities in order to solve the equation. The interval is always going to be \(\textup{angle}\:\theta=0\le \theta \le 2\pi\), which will help when it comes to solving the equation.

Solve the following equation on the interval \(0\le \theta \le 2\pi\): \[\cos \theta=\sin \theta\]

\[\begin{matrix} \cos \theta=\sin \theta\\\\ \frac{\cos \left(\theta\right)}{\cos \left(\theta\right)}=\frac{\sin \left(\theta\right)}{\cos \left(\theta\right)}\\\\ 1=\frac{\sin \left(\theta\right)}{\cos \left(\theta\right)}\\\\ 1=\tan \left(\theta\right)\\ \tan \left(\theta\right)=1\\\\ \tan \left(\frac{\pi}{4}\right)=1\\ \tan \left(\frac{5\pi}{4}\right)=1\\\\ \theta=\frac{\pi}{4},\frac{5\pi}{4}\\\\ \boxed{\mathbf{{\color{red} \left \{ \frac{\pi}{4},\frac{5\pi}{4} \right \} }}} \end{matrix}\]

Solve the following equation on the interval \(0\le \theta \le 2\pi\): \[\sin \theta+\sqrt{3}\cos \theta=-1\]

\[\begin{matrix} \sin \theta+\sqrt{3}\cos \theta=-1\\ \sin \theta=-1-\sqrt{3}\cos \theta\\ (\sin \theta)^2=(-1-\sqrt{3}\cos \theta)^2 \end{matrix}\]\[\begin{matrix} \sin^2 \theta=1+2\sqrt{3}\cos\theta+3\cos^2\theta\\ 1-\cos^2 \theta=1+2\sqrt{3}\cos\theta+3\cos^2\theta\\ 1-\cos^2 \theta-1-2\sqrt{3}\cos \left(\theta\right)-3\cos ^2\left(\theta\right)=0\\ -3\cos ^2\left(\theta\right)-\cos ^2\left(\theta\right)-2\sqrt{3}\cos \left(\theta\right)-1+1=0\\ (-3-1)\cos ^2\left(\theta\right)-2\sqrt{3}\cos \left(\theta\right)+(-1+1)=0\\ (-4)\cos ^2\left(\theta\right)-2\sqrt{3}\cos \left(\theta\right)+(0)=0\\ -4\cos ^2\left(\theta\right)-2\cos \left(\theta\right)\sqrt{3}=0 \end{matrix}\]When it comes to \(u\)-substitution, you first let \(u=\cos\theta\), which means that \(\left\{\begin{matrix} \textup{If}\: u=\cos\theta\\ \textup{Then}\: u^2=\cos^2\theta\\ \end{matrix}\right. .\) with the following being the next set of steps: \[\begin{matrix} &-4\cos ^2\left(\theta\right)-2\cos \left(\theta\right)\sqrt{3}=0&\\ &-4u^2-2u\sqrt{3}=0&\\\\ &u=\frac{-\left(-2\sqrt{3}\right)\pm\sqrt{\left(-2\sqrt{3}\right)^2-4\left(-4\right)(0)}}{2\left(-4\right)}&\\ &u=\frac{2\sqrt{3}\pm\sqrt{12-0}}{(-8)}&\\ &u=-\frac{2\sqrt{3}\pm\sqrt{12}}{8}&\\ &u=-\frac{2\sqrt{3}\pm -2\sqrt{3}}{8}&\\ u=-\frac{2\sqrt{3}+ (-2\sqrt{3})}{8}&&u=-\frac{2\sqrt{3}-(-2\sqrt{3})}{8}\\ u=-\frac{0}{8}&&u=-\frac{2\sqrt{3}+2\sqrt{3}}{8}\\ u=0&&u=-\frac{4\sqrt{3}}{8}\\ &&u=-\frac{\sqrt{3}}{2} \end{matrix}\]

Now that the \(u\)-substitution is done, you have to plug \(\cos\theta\) back in since we let \(u=\cos\theta\) and now find \(\theta\), which means doing the following: \[\begin{matrix} u=0&&u=-\frac{\sqrt{3}}{2}\\\\ \cos\theta=0&&\cos\theta=-\frac{\sqrt{3}}{2}\\\\ \cos\left ( \frac{\pi}{2} \right )=0&&\cos\left ( \frac{5\pi}{6} \right )=-\frac{\sqrt{3}}{2}\\ \cos\left ( \frac{3\pi}{2} \right )=0&&\cos\left ( \frac{7\pi}{6} \right )=-\frac{\sqrt{3}}{2}\\\\ \theta=\frac{\pi}{2},\frac{3\pi}{2}&&\theta=\frac{5\pi}{6},\frac{7\pi}{6}\\\\ &\left \{ \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6} \right \}& \end{matrix}\]

Now, it would be best to check all of the solutions by plugging them back into the original equation to make sure that we have any verified solutions, which can be done by doing the following: \[\begin{matrix} \theta=\frac{\pi}{2} && \theta=\frac{3\pi}{2} \\\\ \sin \left(\frac{\pi }{2}\right)+\sqrt{3}\cos \left(\frac{\pi }{2}\right)=-1&&\sin \left(\frac{3\pi }{2}\right)+\sqrt{3}\cos \left(\frac{3\pi }{2}\right)=-1\\ 1=-1&&-1=-1\\ 1\neq -1&&\textup{True}\\\\

\theta=\frac{5\pi}{6} &&\theta=\frac{7\pi}{6}\\\\ \sin \left(\frac{5\pi}{6}\right)+\sqrt{3}\cos \left(\frac{5\pi}{6}\right)=-1&&\sin \left(\frac{7\pi}{6}\right)+\sqrt{3}\cos \left(\frac{7\pi}{6}\right)=-1\\ -1=-1&&-2=-1\\\\ \textup{True}&&-2\neq -1\\\\

\end{matrix}\]

Based on verifying, the correct answers are \(\boxed{\mathbf{{\color{red} \left \{ \frac{3\pi}{2},\frac{5\pi}{6} \right \} }}}\).

## Section Z. - Find the Equation for the Graph

There are some clever steps as to how to find the equation just by looking at the graph. Based on the ones for this study guide, the functions are only based on the amplitude and change in the period, which would be the following: \[Functions\rightarrow \left\{\begin{matrix} y=A\sin\left(Bx\right)\\y=A\cos\left(Bx\right) \end{matrix}\right.\]

Remember the following that was listed in Section E: \[\begin{matrix} \textup{\textbf{Amplitude}}&\textup{\textbf{Period}}&\textup{\textbf{Phase/Horizontal Shift}}&\textup{\textbf{Vertical Shift}}& \\ \left | a \right |&\frac{2\pi}{b}&c&d& \\ \end{matrix}\]

The amplitude can be determined based on looking at the first curve to the right of the \(y\)-axis. Then look at where the curve will slope to \(0\); that line will be the "\(A\)" in the function. However, when it comes to the change in the period, with the \(x\)-axis being the "crossing line", see how many curves there are from \(0\) to \(\pi\). The number of curves will be the "\(B\)" in the function.

Find an equation for the following graph:

Is this graph going through the origin? Yes. \[\rightarrow y=\sin(x)\]What is the amplitude of this graph? \(5.\) \[\rightarrow y=5\sin(x)\]

How many curves are there between \(0\) and \(\pi\)? \(2.\) \[\rightarrow \boxed{\mathbf{{\color{red} y=5\sin(2x) }}}\]

## Section AA. - Finding Exact Values of Expressions

This is literally the same concepts as Sections F and S. You need to use trivial identities and sometimes use "if-then" scenarios in order to solve these expressions.

Find the exact value of the following expression. Do not use a calculator. \[\sin^2 60^{\circ}-\cos^2 45^{\circ}-\sin^2 30^{\circ}\]

\[\begin{matrix} \textup{\textbf{If}}&\textup{\textbf{Then}}\\ \sin \left(60^{\circ }\right)=\frac{\sqrt{3}}{2}&\sin^2 \left(60^{\circ }\right)=\left (\frac{\sqrt{3}}{2} \right )^2\\ \cos \left(45^{\circ }\right)=\frac{\sqrt{2}}{2}&\cos^2 \left(45^{\circ }\right)=\left (\frac{\sqrt{2}}{2} \right )^2\\ \sin \left(30^{\circ }\right)=\frac{1}{2}&\sin^2 \left(30^{\circ }\right)=\left (\frac{1}{2} \right )^2 \end{matrix}\]\[\begin{matrix} \sin^2 60^{\circ}-\cos^2 45^{\circ}-\sin^2 30^{\circ}\\ =\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{2}}{2}\right)^2-\left(\frac{1}{2}\right)^2\\ =\frac{3}{4}-\frac{2}{4}-\frac{1}{4}\\\\ =\frac{3-2}{4}-\frac{1}{4}\\\\ =\frac{1}{4}-\frac{1}{4}\\\\ =\frac{1-1}{4}\\\\ =\frac{0}{4}\\\\ \boxed{\mathbf{{\color{red} 0 }}} \end{matrix}\]

**Cite as:**Precalculus Mathematics 2 Study Guide.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/precalculus-mathematics-2-study-guide/