Chapter 12.3, Problem 43WE

To determine

Find the exact value of six trigonometric functions of the given angle.

Answer to Problem 43WE

  $\sin \left(-150°\right)=-\frac{1}{2}$

  $\cos \left(-150°\right)=-\frac{\sqrt{3}}{2}$

  $\tan \left(-150°\right)=\frac{\sqrt{3}}{3}$

  $\csc \left(-150°\right)=-2$

  $\sec \left(-150°\right)=-\frac{2\sqrt{3}}{3}$

  $\cot \left(-150°\right)=\sqrt{3}$

Explanation of Solution

Given:

  $\theta =-150°$

Formula Used:

  $\sin \left(180°-\theta \right)=\sin \theta$

  $\cos \left(180°-\theta \right)=-\cos \theta$

  $\cos \left(-\theta \right)=\cos \theta$

  $\sin \left(-\theta \right)=-\sin \theta$

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

  $\theta =-150°=-\left(180°-30°\right)$

  $\begin{array}{l}\sin \left(-150°\right)\\ =\sin \left[-\left(180°-30°\right)\right]\\ =-\sin \left(180°-30°\right)\\ =-\sin 30°\mathrm{..............}[\sin \left(180°-\theta \right)=\sin \theta ]\\ =-\frac{1}{2}\end{array}$

  $\begin{array}{l}\cos \left(-150°\right)\\ =\cos (180°-30°)\mathrm{.........}[\cos \left(-\theta \right)=\cos \theta ]\\ =-\cos 30°\mathrm{.......................}[\cos \left(180°-\theta \right)=-\cos \theta ]\\ =-\frac{\sqrt{3}}{2}\end{array}$

  $\begin{array}{l}\tan \left(-150°\right)\\ =\frac{\sin \left(-150°\right)}{\cos \left(-150°\right)}\\ =\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}\\ =\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\mathrm{...............}[rationalize]\\ =\frac{\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\csc \left(-150°\right)\\ =\frac{1}{\sin \left(-150°\right)}\\ =\frac{1}{\left(-\frac{1}{2}\right)}\\ =-2\end{array}$

  $\begin{array}{l}\sec \left(-150°\right)\\ =\frac{1}{\cos \left(-150°\right)}\\ =\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}\\ =-\frac{2\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\cot \left(-150°\right)\\ =\frac{1}{\tan \left(-150°\right)}\\ =\sqrt{3}\end{array}$