Chapter 6.3, Problem 11E

To determine

To calculate: The value of the trigonometric function $\sin 150°$ .

Answer to Problem 11E

The value of the trigonometric function $\sin 150°$ is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The trigonometric function $\sin 150°$ .

Formula used:

Coordinate plane is divided into four quadrants.

In the first quadrant all trigonometric functions that is $\sin \theta ,cos\theta ,\tan \theta ,\csc \theta ,\sec \theta \text{ and }\cot \theta$ are positive.

In the second quadrant only sine and cosecant trigonometric functions that is $\sin \theta \text{ and }\csc \theta$ are positive and remaining $cos\theta ,\tan \theta ,\sec \theta \text{ and }\cot \theta$ are negative.

In the third quadrant only tangent and cotangent trigonometric functions that is $\tan \theta \text{ and }\cot \theta$ are positive and remaining $cos\theta ,sin\theta ,\sec \theta \text{ and }\csc \theta$ are negative.

In the fourth quadrant only cosine and secant trigonometric functions that is $\cos \theta \text{ and }\sec \theta$ are positive and remaining $\sin \theta ,\tan \theta ,\csc \theta \text{ and }\cot \theta$ are negative.

The reference angle $\overline{\theta }$ is the angle associated with the angle $\theta$ in the standard position. It is the acute angle that is formed by the terminal side of $\theta$ and the x -axis.

Calculation:

Consider the provided trigonometric function $\sin 150°$ .

Denote $\theta =150°$ .

First estimate the reference angle $\overline{\theta }$ associated with the angle $\theta$ .

From the figure provided below,

The provided angle has terminal side in Quadrant $\text{II}$ .

The reference angle is,

  $\begin{array}{c}\overline{\theta }=180°-150°\\ =30°\end{array}$

Recall that coordinate plane is divided into four quadrants.

In the second quadrant only sine and cosecant trigonometric functions that is $\sin \theta \text{ and }\csc \theta$ are positive and remaining $cos\theta ,\tan \theta ,\sec \theta \text{ and }\cot \theta$ are negative.

So, sine trigonometric function, $\sin 150°$ is positive.

Therefore,

  $\begin{array}{c}\sin 150°=\sin 30°\\ =\frac{1}{2}\end{array}$

Thus, the value of the trigonometric function $\sin 150°$ is $\frac{1}{2}$ .