Chapter 5, Problem 3PS

To determine

To verify: that for all integers n ,

  $\sin \left[\frac{\left(12n+1\right)\pi }{6}\right]=\frac{1}{2}$ .

Explanation of Solution

Let $\theta$ be an angle such that $\sin \theta =\frac{1}{2}$

Now, it is known that if $\theta$ lie in the Quadrant I, that is $0\le \theta \le \frac{\pi }{2}$ , then $\sin \theta =\frac{1}{2}$ implies that $\theta =\frac{\pi }{6}$

It is also known that the sine function is a periodic function of the period $2\pi$ .

So, even if any integer multiple of $2\pi$ is added to $\frac{\pi }{6}$ , the equation $\sin \theta =\frac{1}{2}$ is still satisfied.

That is, for any integer n ,

  $\begin{array}{c}\sin \left[\frac{\pi }{6}+2n\pi \right]=\frac{1}{2}\\ \sin \left[\frac{\left(12n+1\right)\pi }{6}\right]=\frac{1}{2}\end{array}$