Chapter 13, Problem 11PT

To determine

To find: Value of $\sin {585}^0$ .

Answer to Problem 11PT

  $\sin {585}^0=-\frac{1}{\sqrt{2}}$

Explanation of Solution

Given information:

  $\sin {585}^0$

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

  $\sin {585}^0$

Here $\theta ={585}^0$

It means

It completes on rotation and extra ${225}^0$ .

Therefore,

Coterminal angle $={585}^0-{360}^0={225}^0$

The terminal side of ${225}^0$ lies in Quadrant − III

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}=\theta -{180}^0\\ \Rightarrow {\theta }^{\text{'}}={225}^0-{180}^0\\ \Rightarrow {\theta }^{\text{'}}={45}^0\end{array}$

Therefore,

  $\sin {585}^0=-\sin {45}^0$

Because the Sine function is negative in Quadrant − III

  $\Rightarrow \sin {585}^0=-\frac{1}{\sqrt{2}}$

Therefore,

  $\sin {585}^0=-\frac{1}{\sqrt{2}}$