Exact Values of Trigonometric Functions – Questions With Answers

Find exact values of trigonometric functions without using a calculator. Questions with full solutions and answers are presented. The trigonometric identities and formulas on this site may be used to solve the questions below.

Let us assume that we want to find the exact value of \( f(x) \), where \( f \) is any of the six trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. To find the exact value of \( f(x) \), we suggest the following steps:

If the angle \( x \) is negative, we first use an identity for negative angles such as \[ \sin(-x) = -\sin(x), \qquad \cos(-x) = \cos(x) \] and similar identities for the other trigonometric functions.
Next, we locate the terminal side of the angle in question, either directly or by using a positive coterminal angle, which allows us to determine the sign of the trigonometric function.
We then find the reference angle \( T_r \) corresponding to the angle in question and use the fact that \[ f(x) = \pm f(T_r) \] where the sign \( + \) or \( - \) is determined by the quadrant found in the previous step. If the angle (or its coterminal angle) lies in Quadrant I, this step is not needed.

Question 1

Find the exact value of \[ \sin\!\left(-\frac{\pi}{3}\right). \]

Solution to Question 1:

Use the identity for negative angles to write \[ \sin\!\left(-\frac{\pi}{3}\right) = -\sin\!\left(\frac{\pi}{3}\right). \]
The angle \( \frac{\pi}{3} \) lies in Quadrant I, so there is no need to use either a coterminal angle or a reference angle. The value is evaluated directly: \[ \sin\!\left(-\frac{\pi}{3}\right) = -\sin\!\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}. \]

Question 2

Find the exact value of \[ \cos(-390^\circ). \]

Solution to Question 2:

We use the identity \( \cos(-x) = \cos(x) \) to write \[ \cos(-390^\circ) = \cos(390^\circ). \]
Since \( 390^\circ \) is greater than \( 360^\circ \), we find a positive coterminal angle \( t \), less than \( 360^\circ \), as follows: \[ t = 390^\circ - 360^\circ = 30^\circ. \]
Because \( 390^\circ \) and \( 30^\circ \) are coterminal angles, we can write \[ \cos(390^\circ) = \cos(30^\circ). \]
Finally, \[ \cos(-390^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2}. \] There is no need to use a reference angle since \( 30^\circ \) lies in Quadrant I.

Question 3

Find the exact value of \[ \sec\!\left(\frac{3\pi}{4}\right). \]

Solution to Question 3:

The angle \( \frac{3\pi}{4} \) has its terminal side in Quadrant II. In Quadrant II, the secant function is negative. Hence, \[ \sec\!\left(\frac{3\pi}{4}\right) = -\sec(T_r). \]
The reference angle \( T_r \) corresponding to \( \frac{3\pi}{4} \) is \[ T_r = \pi - \frac{3\pi}{4} = \frac{\pi}{4}. \]
Therefore, \[ \sec\!\left(\frac{3\pi}{4}\right) = -\sec\!\left(\frac{\pi}{4}\right) = -\sqrt{2}. \]

Question 4

Find the exact value of \[ \cot(840^\circ). \]

Solution to Question 4:

The angle \( 840^\circ \) is positive and greater than \( 360^\circ \), so we first find a coterminal angle: \[ t = 840^\circ - 2(360^\circ) = 120^\circ. \]
We can therefore write \[ \cot(840^\circ) = \cot(120^\circ). \]
The angle \( 120^\circ \) lies in Quadrant II, where the cotangent function is negative. Hence, \[ \cot(120^\circ) = -\cot(T_r). \]
The reference angle corresponding to \( 120^\circ \) is \[ T_r = 180^\circ - 120^\circ = 60^\circ. \]
Finally, \[ \cot(840^\circ) = -\cot(60^\circ) = -\frac{\sqrt{3}}{3}. \]

Question 5

Find the exact value of \[ \csc\!\left(-\frac{7\pi}{4}\right). \]

Solution to Question 5:

Using the negative-angle identity, we have \[ \csc\!\left(-\frac{7\pi}{4}\right) = -\csc\!\left(\frac{7\pi}{4}\right). \]
The terminal side of \( \frac{7\pi}{4} \) lies in Quadrant IV, where the cosecant function is negative. The reference angle is \[ T_r = 2\pi - \frac{7\pi}{4} = \frac{\pi}{4}. \]
Hence, \[ \csc\!\left(\frac{7\pi}{4}\right) = -\csc\!\left(\frac{\pi}{4}\right) = -\sqrt{2}. \]
Substituting this into the negative-angle identity gives \[ \csc\!\left(-\frac{7\pi}{4}\right) = \sqrt{2}. \]

Question 6

Find the exact value of \[ \cot\!\left(\frac{121\pi}{3}\right). \]

Solution to Question 6:

First note that \[ \frac{121\pi}{3} = \frac{120\pi}{3} + \frac{\pi}{3} = 40\pi + \frac{\pi}{3}. \]
A positive coterminal angle is obtained by subtracting multiples of \( 2\pi \): \[ \frac{121\pi}{3} - 20(2\pi) = \frac{\pi}{3}. \]
The coterminal angle \( \frac{\pi}{3} \) lies in Quadrant I, so no reference angle is required: \[ \cot\!\left(\frac{121\pi}{3}\right) = \cot\!\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{3}. \]

Question 7

Find the exact value of \[ \sec(-3810^\circ). \]

Solution to Question 7:

Using the negative-angle identity: \[ \sec(-3810^\circ) = \sec(3810^\circ). \]
Note that \[ 3810^\circ = 3600^\circ + 210^\circ. \]
A coterminal angle is therefore \[ t = 3810^\circ - 10(360^\circ) = 210^\circ. \]
The angle \( 210^\circ \) lies in Quadrant III, where the secant function is negative. Hence, \[ \sec(3810^\circ) = -\sec(T_r). \]
The reference angle is \[ T_r = 210^\circ - 180^\circ = 30^\circ. \]
Therefore, \[ \sec(-3810^\circ) = -\sec(30^\circ) = -\frac{2}{\sqrt{3}}. \]