Fundamental Trigonometric Identities - Problem Solving (Easy)

When working with trigonometric identities, it may be useful to keep the following tips in mind:

• Draw a picture illustrating the problem if it involves only the basic trigonometric functions.
• If the problem expresses an identity between trigonometric functions, try working on one side of the identity to write the trigonometric functions from one side in terms of trigonometric functions on the other side.
• Use the Pythagorean identities to compute values in the trigonometric identity.
• Multiply rational expressions by conjugates in order to take advantage of the Pythagorean identities.
• Add rational expressions by finding common denominators.

The fundamental period of the graphs of \(\sin x, \cos x, \csc x, \sec x \) is \(2\pi,\) while the fundamental period of the graphs of \( \tan x, \cot x \) is \(\pi\).

Pythagorean identities:

\[\begin{eqnarray} \sin^2 A + \cos^2 A &=& 1 \\ \tan^2 A + 1 &=& \sec^2 A \\ \cot^2 A + 1 &=& \csc^2 A. \end{eqnarray} \]

Compound angle formulas:

\[\begin{eqnarray} \sin(A\pm B) &=& \sin A \cos B \pm \cos A \sin B \\ \cos(A\pm B) &=& \cos A \cos B \mp \sin A \sin B \\ \tan(A\pm B) &=& \frac{ \tan A \pm \tan B } { 1 \mp \tan A \tan B }. \end{eqnarray} \]

You may find the following table useful:

\[ \begin{array} {| c | c | c | c | c | c |} \hline \theta & 0^\circ & \frac{\pi}{6} = 30^\circ & \frac{\pi}{4} = 45^\circ & \frac{\pi}{3} = 60^\circ & \frac{\pi}{2} = 90^\circ\\ \hline \sin \theta & \frac {\sqrt{0}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{4}} {2} \\ \hline \cos \theta & \frac {\sqrt{4}} {2} & \frac {\sqrt{3}} {2} & \frac {\sqrt{2}} {2} & \frac {\sqrt{1}} {2} & \frac {\sqrt{0}} {2}\\ \hline \tan \theta & 0 & \frac { 1}{\sqrt{3} } & 1 & \sqrt{3} & \infty \\ \hline \end{array}\]

For any angle \(\theta,\) we have

\[\begin{align} \cos^2 \theta + \sin^2 \theta &= 1 \\ 1 + \tan^2 \theta &= \sec^2 \theta \\ \cot^2 \theta + 1 &= \csc^2 \theta. \end{align}\]

For any angle \(\theta,\) we have

\[\begin{align} \cos(-\theta) &= \cos \theta \\ \sin(-\theta) &= -\sin \theta\\ \tan(-\theta) &= -\tan \theta\\ \cot(-\theta) &= -\cot \theta\\ \csc(-\theta) &= -\csc \theta\\ \sec(-\theta) &= \sec \theta. \end{align}\]

For any angle \(\theta,\) we have

\[\begin{align} \sin \theta &=\sin(\theta+2\pi) &\quad \csc \theta &=\csc(\theta+2\pi)\\ \cos \theta &=\cos(\theta+2\pi) &\quad \sec \theta &=\sec(\theta+2\pi)\\ \tan \theta &=\tan(\theta+\pi) &\quad \cot \theta &=\cot(\theta+\pi). \end{align}\]

For any angle \(\theta,\) we have

\[ \begin{align} \cos \theta & = \sin \left( \frac{\pi}{2} - \theta \right) \\ \sin \theta &= \cos \left( \frac{\pi}{2}-\theta \right) \\ \cot \theta & = \tan \left( \frac{\pi}{2} - \theta \right) \\ \csc \theta & = \sec \left( \frac{\pi}{2} - \theta \right). \end{align}\]

For any angle \(\theta,\) we have

\[\begin{align} \sin 2 \theta &= 2\sin \theta \cos \theta \\\\ \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= 2\cos^2 \theta - 1\\ &= 1 - 2\sin^2 \theta \\\\ \tan 2\theta &= \frac{2\tan \theta}{1 - \tan^2 \theta}. \end{align} \]

For any angle \(\theta,\) we have

\[\begin{align} \sin(x+y) &= \sin x \cos y + \cos x \sin y \\ \sin(x-y) &= \sin x \cos y - \cos x \sin y \\\\ \cos(x+y) &= \cos x \cos y − \sin x \sin y \\ \cos(x-y) &= \cos x \cos y + \sin x \sin y \\\\ \tan(x+y) &= \dfrac{\tan x + \tan y}{1 - \tan x \tan y} \\ \tan(x-y) &= \dfrac{\tan x - \tan y}{1 + \tan x \tan y}. \end{align} \]

For any angle \(\theta,\) we have

\[\begin{align} \sin 3 \theta &= 3 \sin \theta - 4 \sin ^3 \theta \\ \cos 3\theta &= 4 \cos ^ 3 \theta - 3 \cos \theta. \end{align} \]

For any angle \(\theta,\) we have

\[\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}} \\ \cos\theta=\frac{1-\tan^{2}\frac{\theta}{2}}{1+\tan^{2}\frac{\theta}{2}} \\ \tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^{2}\frac{\theta}{2}}.\]

For any angle \(\theta,\) we have

\[\begin{align} \sin^2 \theta &=\frac{1}{2}(1-\cos 2\theta) \\ \cos^2 \theta &=\frac{1}{2}(1+\cos 2\theta). \end{align} \]

For any angle \(\theta,\) we have

\[\begin{align} \cos x \cos y & = \frac{1}{2} \big(\cos (x - y) + \cos(x+ y) \big) \\ \sin x \cos y &= \frac{1}{2} \big(\sin (x- y) + \sin(x + y) \big) \\ \cos x \sin y &= \frac{1}{2} \big(\sin (x + y) - \sin(x - y) \big)\\ \sin x \sin y &= \frac{1}{2} \big(\cos (x - y) - \cos(x + y) \big). \end{align}\]

For any angle \(\theta,\) we have

\[\begin{align} \sin x+\sin y & =2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) \\ \cos x+\cos y &=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right). \\ \end{align}\]

If \(\sin x = \cos x = 0.2\), what is the value of \(\cos 2x?\)

ANSWER

Because the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) is not fulfilled, there is no solution of \(x\) and thus there is no possible value for \(\cos 2x \). \(_\square\)

If \(\sin \theta = \frac45\) and \(\theta\) is not in the first quadrant, then find the value of \(\cos \theta\).

ANSWER

Sine is positive only in the first and second quadrants but \(\theta\) is not in the first quadrant, so definitely \(\theta\) must lie in the second quadrant where \(\cos \theta < 0:\)

\[\begin{align} \sin^2 \theta + \cos^2 \theta & = 1 \\ \cos^2 \theta & = 1 - \sin^2 \theta \\ & = 1 - \dfrac{16}{25} = \dfrac{9}{25} \\ \Rightarrow \cos \theta & = \pm \dfrac35 \\ \Rightarrow \cos \theta & = - \dfrac35.\ _\square \qquad (\text{since } \cos \theta < 0) \\ \end{align}\]

If \(\sec \theta + \tan \theta = \frac23\), find the value of \(\sin \theta\) and determine the quadrant in which \(\theta\) lies.

ANSWER

We know that \(\sec^2 \theta - \tan^2 \theta = 1 \implies \sec \theta - \tan \theta - \frac{1}{\sec \theta + \tan \theta} = \dfrac{1}{\hspace{2mm} \frac23\hspace{2mm} } = \frac32\).

By adding, \((\sec \theta + \tan \theta) + (\sec \theta - \tan \theta) = \frac23 + \frac32 \implies 2 \sec \theta = \frac{13}{6} \implies \sec \theta = \frac{13}{12}\).

By subtracting, \((\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = \frac23 - \frac32 \implies 2 \tan \theta = \frac{-5}{6} \implies \tan \theta = \frac{-5}{12}\).

From this we get to know that \(\sec \theta\) is positive and \(\tan \theta\) is negative. This is only possible when \(\theta\) lies in the fourth quadrant.

Therefore, \( \sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{-5}{13}.\) \(_\square\)

Find the value of \(2\big(\sin^6 \theta + \cos^6 \theta\big) - 3\big(\sin^4 \theta + \cos^4 \theta\big)\).

ANSWER

We have

\[\begin{align} 2\big(\sin^6 \theta + \cos^6 \theta\big) - 3\big(\sin^4 \theta + \cos^4 \theta\big) & = 2\left[\big(\sin^2 \theta\big)^3 + \big(\cos^2 \theta\big)^3\right] - 3\left[\big(\sin^2 \theta\big)^2 + \big(\cos^2 \theta\big)^2\right] \\ & = 2\left[\big(\sin^2 \theta + \cos^2 \theta\big)^3 - 3\sin^2 \theta \cos^2 \theta\big(\sin^2 \theta + \cos^2 \theta\big)\right] - 3\left[\big(\sin^2 \theta + \cos^2 \theta\big)^2 - \sin^2 \theta \cos^2 \theta\right] \\ & = 2\big(1 - 3\sin^2 \theta \cos^2 \theta\big) - 3\big(1 - 2\sin^2 \theta \cos^2 \theta\big) \\ & = 2 - 3 \\ & = -1.\ _\square \end{align}\]

Find the value of

1. \(\,(\tan \theta + \cot \theta)^2 \)
2. \(\ \sec^2 \theta + \csc^2 \theta.\)

ANSWER

We have

\(\begin{align} 1.\ (\tan \theta + \cot \theta)^2 & = \tan^2 \theta + \cot^2 \theta + 2\tan \theta \cot \theta \\ & = \tan^2 \theta + \cot^2 \theta + 2 \\ & = \big(1 + \tan^2 \theta\big) + \big(1 + \cot^2 \theta\big) \\ & = \sec^2 \theta + \csc^2 \theta \end{align}\)

\(\begin{align}
2.\ \sec^2 \theta + \csc^2 \theta & = \dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta} \\ \\ & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cdot \cos^2 \theta} \\ \\ & = \dfrac{1}{\sin^2 \theta \cdot \cos^2 \theta} \\ \\ & = \sec^2 \theta \cdot \csc^2 \theta.\ _\square \end{align}\)

In this summary, we collect all of the trigonometric identities that are useful to know for problem-solving.

Find the value of \(\sin^2 \frac{\pi}{10} + \sin^2 \frac{4\pi}{10} + \sin^2 \frac{6 \pi}{10} + \sin^2 \frac{9 \pi}{10}\).

ANSWER

We have

\[\begin{align} \sin^2 \dfrac{\pi}{10} + \sin^2 \dfrac{4\pi}{10} + \sin^2 \dfrac{6 \pi}{10} + \sin^2 \dfrac{9 \pi}{10} &=\sin^2 \left(\dfrac{\pi}{10}\right) + \sin^2 \left( \dfrac{\pi}{2} - \dfrac{\pi}{10} \right) + \sin^2 \left(\dfrac{\pi}{2} + \dfrac{\pi}{10} \right) + \sin^2 \left(\pi - \dfrac{\pi}{10} \right) \\ &=\underbrace{\sin^2 \dfrac{\pi}{10} + \cos^2 \dfrac{\pi}{10}}_{1} + \underbrace{\cos^2 \dfrac{\pi}{10} + \sin^2 \dfrac{\pi}{10}}_{1} \\ &=1 + 1 \\ &= 2.\ _\square \end{align}\]

Find the value of \(\cot \frac{\pi}{16} \times \cot \frac{2 \pi}{16} \times \cot \frac{3 \pi}{16} \times \cdots \times \cot \frac{7 \pi}{16}\).

ANSWER

We have

\[\begin{align} \cot \dfrac{\pi}{16} \cdot \cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{3 \pi}{16} \times \cdots \times \cot \dfrac{7 \pi}{16} &=\left(\cot \dfrac{\pi}{16} \cdot \cot \dfrac{7 \pi}{16}\right) \cdot \left(\cot \dfrac{2 \pi}{16} \cdot \cot \dfrac{6 \pi}{16} \right) \cdot \left(\cot \dfrac{3 \pi}{16} \cdot \cot \dfrac{5 \pi}{16} \right) \cdot \cot \dfrac{4 \pi}{16} \\ &=\left(\cot \dfrac{\pi}{16} \cdot \cot \Big(\dfrac{\pi}{2} - \dfrac{\pi}{16}\Big)\right) \cdot \left(\cot \dfrac{2 \pi}{16} \cdot \cot \Big(\dfrac{\pi}{2} - \dfrac{2\pi}{16} \Big)\right) \cdot \left(\cot \dfrac{3 \pi}{16} \cdot \cot \Big(\dfrac{\pi}{2} - \dfrac{3\pi}{16} \Big)\right) \cdot \cot \dfrac{\pi}{4} \\ &=\underbrace{\left(\cot \dfrac{\pi}{16} \cdot \tan \dfrac{\pi}{16}\right)}_{1} \cdot \underbrace{\left(\cot \dfrac{2 \pi}{16} \cdot \tan \dfrac{2\pi}{16} \right)}_{1} \cdot \underbrace{\left(\cot \dfrac{3 \pi}{16} \cdot \tan \dfrac{3\pi}{16} \right)}_{1} \cdot 1 \\ &= 1.\ _\square \end{align}\]

If \(3 \sin \theta + 4 \cos \theta = 5\), then find the value of \(4 \sin \theta - 3 \cos \theta\).

ANSWER

Given \(3 \sin \theta + 4 \cos \theta = 5,\) let \(4\sin \theta - 3\cos \theta = a\). On squaring and adding both these equations, we get

\[\begin{align} 5^2 & = (3\sin \theta + 4 \cos \theta)^2 \\ a^2 & = (4\sin \theta - 3\cos \theta)^2 \\ \\ 5^2 + a^2 & = (3\sin \theta + 4 \cos \theta)^2 + (4\sin \theta - 3\cos \theta)^2 \\ a^2 + 25 & = 9\sin^2 \theta + 16\cos^2 \theta + 24\sin\theta\cos\theta + 16\sin^2\theta + 9\cos^2\theta - 24\sin\theta\cos\theta \\ a^2 + 25 & = 25(\sin^2 \theta + \cos^2 \theta) \\ a^2 + 25 & = 25 \\ a & = 0 \\ \\ \Rightarrow 4\sin\theta - 3\cos\theta & = 0.\ _\square \end{align}\]

If \(\cos \theta + \sin \theta = \sqrt{2} \cos \theta\), find the value of \(\cos \theta - \sin \theta\).

ANSWER

Given \(\cos \theta + \sin \theta = \sqrt{2}\cos \theta,\) we have \(\big(\sqrt{2} - 1\big)\cos\theta = \sin\theta\). On multiplying both the sides with \(\big(\sqrt2 + 1\big),\) we get

\[\begin{align} \big(\sqrt2 + 1\big)\big(\sqrt2 - 1\big)\cos\theta & = \big(\sqrt2 + 1\big)\sin \theta \\ (2 - 1)\cos\theta & = \sqrt2 \sin\theta + \sin\theta \\ \\ \Rightarrow \cos\theta - \sin\theta & = \sqrt2 \sin\theta.\ _\square \end{align}\]