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.

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: $\sin\:22.5^{\circ}$

Reveal the answer

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: $\begin{matrix} \sqrt{\frac{1-\cos \left(45^{\circ }\right)}{2}}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{\frac{2}{2}-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}=\boxed{\mathbf{{\color{#D61F06} \frac{1}{2}\sqrt{2-\sqrt{2}} }}} \end{matrix}$

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: $\cos\:165^{\circ}$

Reveal the answer

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: $\begin{matrix} \sqrt{\frac{1-\sin \left(240^{\circ }\right)}{2}}\\ =-\sqrt{\frac{1-\left(-\frac{\sqrt{3}}{2}\right)}{2}}\\ =-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{\frac{2}{2}+\frac{\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}\\ =-\sqrt{\frac{2+\sqrt{3}}{4}}\\ =-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\\ =-\frac{\sqrt{2+\sqrt{3}}}{2}\\ =\boxed{\mathbf{{\color{#D61F06} -\frac{1}{2}\sqrt{2+\sqrt{3}} }}} \end{matrix}$

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

Reveal the answer

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: $\begin{matrix} \sin \left(x+y\right)&=\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)\\ -\sin \left(x+y\right)&=-[\sin \left(x\right)\cos \left(y\right)+\cos \left(x\right)\sin \left(y\right)]\\ -\sin \left(30^{\circ }+45^{\circ }\right)&=-[\sin \left(30^{\circ }\right)\cos \left(45^{\circ }\right)+\cos \left(30^{\circ }\right)\sin \left(45^{\circ }\right)]\\ &=-\left [\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\right ]\\ &=-\left [\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right ]\\ &=-\left [\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}\right ]\\ &=-\left [\frac{1+\sqrt{3}}{2\sqrt{2}}\right ]\\ &=-\left [ \frac{\left(1+\sqrt{3}\right)\cdot \sqrt{2}}{2\sqrt{2}\cdot \sqrt{2}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{2\sqrt{4}}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{2(2)}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} \right ]\\ &=-\left [ \frac{\sqrt{2+\sqrt{3}}}{2} \right ]\\ &=-\left [ \frac{1}{2}\sqrt{2+\sqrt{3}} \right ]\\ &=\boxed{\mathbf{{\color{#D61F06} -\frac{1}{2}\sqrt{2+\sqrt{3}} }}} \end{matrix}$

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? $\boxed{\textup{P.S. You would have to use Soh-Cah-Toa in order to solve this problem.}}$

Reveal the answer

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: $\begin{matrix} \tan 30^{\circ} =\frac{x}{150}\\ 150\tan 30^{\circ} =x\\ 150\left ( \frac{1}{\sqrt{3}} \right ) =x\\ \frac{150}{\sqrt{3}} =x\\ \frac{150}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} =x\\ \frac{150\sqrt{3}}{3} =x\\ \frac{50\sqrt{3}}{1} =x\\ 50\sqrt{3} =x\\ 86.60254 \approx x\\ 87 \approx x\\ \boxed{\mathbf{{\color{#D61F06} x\approx 87\: feet }}} \end{matrix}$

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. $\boxed{\textup{P.S. You would have to use the law of sines in order to solve this problem.}}$

Reveal the answer

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: $\begin{matrix} \frac{a}{\sin A}=\frac{c}{\sin C}\\\\ \frac{175}{\sin 30^{\circ}}=\frac{x}{\sin 90^{\circ}}\\ 175\sin 90^{\circ}=x\sin 30^{\circ}\\ 175(1)=x\left ( \frac{1}{2} \right )\\ 175=\frac{x}{2}\\ 175\cdot 2=\frac{x}{2}\cdot 2\\ 350=x\\ \boxed{\mathbf{{\color{#D61F06} x=350\: feet }}} \end{matrix}$

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: $\left(7,\frac{2\pi }{3}\right)$

Reveal the answer

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: $\begin{matrix} \left(7,\frac{2\pi }{3}\right)\\ (x,y)\\ (r,\theta)\\\\ (r \cos \theta,r \sin \theta)\\ (7 \cos \frac{2\pi }{3},7 \sin \frac{2\pi }{3})\\ (7 \left ( -\frac{1}{2} \right ),7 \left ( \frac{\sqrt{3}}{2} \right ))\\ \boxed{\mathbf{{\color{#D61F06} \left(-\frac{7}{2},\frac{7\sqrt{3}}{2} \right) }}} \end{matrix}$

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: $\begin{matrix} \sin(\theta)=\frac{8}{17},&0\le \theta \le \frac{\pi }{2}&&&\textup{Find} \cos(2\theta). \end{matrix}$

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

Reveal the answer

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: $\begin{matrix} \cos(2\theta)&=1-2\sin^2\theta\\ &=1-2(\frac{8}{17})^2\\\\ &=1-2(\frac{64}{289})\\\\ &=1-\frac{128}{289}\\\\ &=\frac{289}{289}-\frac{128}{289}\\\\ &=\frac{289-128}{289}\\\\ &=\boxed{\mathbf{{\color{#D61F06} \frac{161}{289} }}} \end{matrix}$

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: $\begin{matrix} \tan(\theta)=\frac{12}{5},&\pi\le \theta \le \frac{3\pi }{2}&&&\textup{Find} \sin(2\theta). \end{matrix}$

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

Reveal the answer

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: $\begin{matrix} \sin(2\theta)&=2\sin\theta \cos\theta\\ &=2(\frac{o}{h}) (\frac{a}{h})\\\\ &=2(\frac{12}{13}) (\frac{5}{13})\\\\ &=2(\frac{60}{169})\\\\ &=\boxed{\mathbf{{\color{#D61F06} \frac{120}{169} }}} \end{matrix}$

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. $\begin{matrix} y=-3\sin x&&\textup{Find the amplitude.} \end{matrix}$

Reveal the answer

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 $\left | a \right |=\left | -3 \right |=\boxed{\mathbf{{\color{#D61F06} 3 }}}$

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

Reveal the answer

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 $\frac{2\pi}{b}=\boxed{\mathbf{{\color{#D61F06} \frac{2\pi}{5} }}}$

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. $\cos 60^{\circ}+\tan 60^{\circ}$

Reveal the answer

$\begin{matrix} \cos\:60^{\circ }+\tan\:60^{\circ }\\ =\frac{1}{2}+\sqrt{3}\\ =\frac{1}{2}+\frac{2\sqrt{3}}{2}\\ =\boxed{\mathbf{{\color{#D61F06} \frac{1+2\sqrt{3}}{2} }}} \end{matrix}$

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.

Reveal the answer

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: $\begin{matrix} A&=\frac{1}{2}ab\sin C\\ &=\frac{1}{2}(6)(8)\sin (110^{\circ})\\ &=\frac{1}{2}(48)\sin (110^{\circ})\\ &=24\sin (110^{\circ})\\ &\boxed{\mathbf{{\color{#D61F06} \approx 22.55\:u^2 }}} \end{matrix}$

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 $(r,\theta)$: $x^2+y^2-4x=0$

Reveal the answer

$\begin{matrix} x^2+y^2-4x=0\\ (x^2+y^2)-4(x)=0\\ (r^2)-4(r \cos\theta)=0\\ r^2-4r \cos\theta=0\\ r^2=4r \cos\theta\\ \frac{r^2}{r}=\frac{4r \cos\theta}{r}\\ \boxed{\mathbf{{\color{#D61F06} r=4 \cos\theta }}} \end{matrix}$

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 $(x,y)$: $r=\frac{5}{1+\cos \left(\theta\right)}$

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$\begin{matrix} r=\frac{5}{1+\cos\theta}\\ r(1+\cos\theta)=5\\ r+r\cos\theta=5\\ r+(r\cos\theta)=5\\ r+(x)=5\\ r+x=5\\ r=5-x\\ (r)^2=(5-x)^2\\ r^2=x^2-10x+25\\ (r^2)=x^2-10x+25\\ (x^2+y^2)=x^2-10x+25\\ x^2+y^2=x^2-10x+25\\ y^2=-10x+25\\ \boxed{\mathbf{{\color{#D61F06} y^2=25-10x }}} \end{matrix}$

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: $\sin(8\theta)\cos(4\theta)$

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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: $\begin{matrix} \sin\left(a\right)\cos\left(b\right)&=\frac{1}{2}\left[\sin\left(a+b\right)+\sin\left(a-b\right)\right]\\ \sin\left(8\theta \right)\cos\left(4\theta \right)&=\frac{1}{2}\left[\sin\left(8\theta +4\theta \right)+\sin\left(8\theta -4\theta \right)\right]\\ &=\boxed{\mathbf{{\color{#D61F06} \frac{1}{2}\left[\sin\left(12\theta \right)+\sin\left(4\theta \right)\right] }}} \end{matrix}$

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: $\sin(7\theta)+\sin(3\theta)$

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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: $\begin{matrix} \sin\left(a\right)+\sin\left(b\right)&=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)\\\\ \sin\left(7\theta \right)+\sin\left(3\theta \right)&=2\sin\left(\frac{7\theta +3\theta }{2}\right)\cos\left(\frac{7\theta -3\theta }{2}\right)\\\\ &=2\sin\left(\frac{10\theta }{2}\right)\cos\left(\frac{4\theta }{2}\right)\\\\ &=\boxed{\mathbf{{\color{#D61F06} 2\sin\left(5\theta \right)+\cos\left(2\theta \right) }}} \end{matrix}$

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. $\textup{Find} \sin A\: \textup{when}\: a=5\: \textup{and}\: b=8.$

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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: $\sin\left(A \right)=\frac{ac}{a^2+b^2}=\frac{5\sqrt{89}}{5^2+8^2}=\frac{5\sqrt{89}}{25+64}=\boxed{\mathbf{{\color{#D61F06} \frac{5\sqrt{89}}{89} }}}$

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 $\theta$: $\begin{matrix} \sin(\theta)=\frac{3\sqrt{10}}{10}&&\textup{Find} \tan(\theta). \end{matrix}$

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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: $\tan\theta=\frac{o}{a}=\frac{3\sqrt{10}}{\sqrt{10}}=\frac{3(1)}{(1)}=\boxed{\mathbf{{\color{#D61F06} 3 }}}$

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: $\sin^{-1}\frac{\sqrt{2}}{2}$

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$\sin^{-1}\frac{\sqrt{2}}{2}$ 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 $\boxed{\mathbf{{\color{#D61F06} \frac{\pi}{4} }}}.$

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. $\begin{matrix} \sin(\theta)>0&\textup{and}&\cos(\theta)<0 \end{matrix}$

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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 $\boxed{\mathbf{{\color{#D61F06} Quadrant\: II }}}$.

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:

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First off, we need to find $\measuredangle B$, which can be done by doing the following: $\begin{matrix} \measuredangle a+\measuredangle B+\measuredangle C=180^{\circ}\\ 50^{\circ}+\measuredangle B+95^{\circ}=180^{\circ}\\ \measuredangle B+145^{\circ}=180^{\circ}\\ \boxed{\mathbf{{\color{#D61F06} \measuredangle B=35^{\circ} }}} \end{matrix}$

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: $\begin{matrix} \textup{side a}&\textup{side c}\\ \frac{a}{\sin A}=\frac{b}{\sin B}&\frac{b}{\sin B}=\frac{c}{\sin C}\\\\ \frac{a}{\sin 50^{\circ}}=\frac{6}{\sin 35^{\circ}}&\frac{6}{\sin 35^{\circ}}=\frac{c}{\sin 95^{\circ}}\\\\ a\sin 35^{\circ}=6\sin 50^{\circ}&6\sin 95^{\circ}=c\sin 35^{\circ}\\ a=\frac{6\sin 50^{\circ}}{\sin 35^{\circ}}&\frac{6\sin 95^{\circ}}{\sin 35^{\circ}}=c\\ a=8.013346378&10.42087473=c\\ a=8.013346378&c=10.42087473\\ \boxed{\mathbf{{\color{#D61F06} a\approx 8.01 }}}&\boxed{\mathbf{{\color{#D61F06} c\approx 10.42 }}} \end{matrix}$