Chapter 7.3, Problem 2AYU

To determine

To solve: To find value of trigonometric expression $\langle \theta =2n\pi +\frac{\pi }{6},\theta =2n\pi +\frac{5\pi }{6}\rangle$ as mentioned in the problem.

Answer to Problem 2AYU

The value of an algebraic expression $\sin \left(\frac{\pi }{4}\right),\cos \left(\frac{8\pi }{3}\right)$ is $\frac{1}{\sqrt{2}},-\frac{1}{2}$ respectively.

Explanation of Solution

Given information:

The mentioned trigonometric expression is $\sin \left(\frac{\pi }{4}\right),\cos \left(\frac{8\pi }{3}\right)$ .

Formula used:

For sine function, we use unit circle.

For cosine function, $\cos \left(2\pi +\theta \right)=\cos \theta ;\cos \left(\pi -\theta \right)=-\cos \theta$ .

Calculation:

Consider the mentioned trigonometric expression $\sin \left(\frac{\pi }{4}\right),\cos \left(\frac{8\pi }{3}\right)$ .

Recall that for sine function, in the first quadrant of the unit circle,

Let $\sin \left(\frac{\pi }{4}\right)=\theta \Rightarrow \frac{\pi }{4}={\sin }^{-1}\theta \Rightarrow {45}^0={\sin }^{-1}\theta \Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$ .

Recall that formula for cosine function is $\cos \left(2\pi +\theta \right)=\cos \theta$ .

Therefore $\cos \left(\frac{8\pi }{3}\right)=\cos \left(2\pi +\frac{2\pi }{3}\right)=\cos \left(\frac{2\pi }{3}\right)$ .

Also, formula for cosine function $\cos \left(\pi +\theta \right)=-\cos \theta$

Therefore $\cos \left(\frac{2\pi }{3}\right)=\cos \left(\pi +\frac{\pi }{3}\right)=-\cos \frac{\pi }{3}=-\frac{1}{2}$ .

So, the value of trigonometric function $\sin \left(\frac{\pi }{4}\right),\cos \left(\frac{8\pi }{3}\right)$ is $\frac{1}{\sqrt{2}},-\frac{1}{2}$ respectively.