Chapter 12, Problem 40SGR

To determine

To find: Value of $\cos \left(-{210}^0\right)$ .

Answer to Problem 40SGR

  $\cos \left(-{210}^0\right)=-\frac{\sqrt{3}}{2}$

Explanation of Solution

Given information:

  $\cos \left(-{210}^0\right)$

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

  $\cos \left(-{210}^0\right)$

Here $\theta =-{210}^0$

Coterminal angle $={360}^0-{210}^0={150}^0$

It means

The terminal side of ${150}^0$ lies in Quadrant − II

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={180}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={180}^0-{150}^0\\ \Rightarrow {\theta }^{\text{'}}={30}^0\end{array}$

Therefore,

  $\cos \left(-{210}^0\right)=-\cos \left({30}^0\right)$

Because the Cos function is negative in Quadrant − II

  $\Rightarrow \cos \left(-{210}^0\right)=-\frac{\sqrt{3}}{2}$

Therefore,

  $\cos \left(-{210}^0\right)=-\frac{\sqrt{3}}{2}$