Chapter 12, Problem 25SGR

To determine

To find: Value of $\cos \frac{3\pi }{2}$ .

Answer to Problem 25SGR

  $\cos \frac{3\pi }{2}=0$

Explanation of Solution

Given information:

  $\cos \frac{3\pi }{2}$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $\cos \frac{3\pi }{2}$

The terminal side of $\frac{3\pi }{2}$ lies in Quadrant − III

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}=\theta -\pi \\ \Rightarrow {\theta }^{\text{'}}=\frac{3\pi }{2}-\pi \\ \Rightarrow {\theta }^{\text{'}}=\frac{3\pi -2\pi }{2}\\ \Rightarrow {\theta }^{\text{'}}=\frac{\pi }{2}\end{array}$

Therefore,

  $\cos \frac{3\pi }{2}=-\cos \frac{\pi }{2}$

Because Cosine function is negative in Quadrant − III

  $\begin{array}{l}\Rightarrow \cos \frac{3\pi }{2}=-0\\ \Rightarrow \cos \frac{3\pi }{2}=0\end{array}$

Therefore,

  $\cos \frac{3\pi }{2}=0$