Chapter 12.3, Problem 39WE

To determine

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

Answer to Problem 39WE

  $\sin 135°=\frac{\sqrt{2}}{2}$

  $\cos 135°=-\frac{\sqrt{2}}{2}$

  $\tan 135°=-1$

  $\csc 135°=\sqrt{2}$

  $\sec 135°=-\sqrt{2}$

  $\cot 135°=-1$

Explanation of Solution

Given:

  $\theta =135°$

Formula Used:

  $\sin \left(180°\pm \theta \right)=\mp \sin \theta$

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

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

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

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

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

Calculation:

  $\theta =135°=180°-45°$

  $\begin{array}{l}\sin 135°\\ =\sin \left(180°-45°\right)\\ =\sin 45°\mathrm{......................}[\sin \left(180°-\theta \right)=\sin \theta ]\\ =\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}\mathrm{...................}[rationalize]\\ =\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}\cos 135°\\ =\cos \left(180°-45°\right)\\ =-\cos 45°\mathrm{.......................}[\cos \left(180°-\theta \right)=\cos \theta ]\\ =-\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}\tan 135°\\ =\frac{\sin 135°}{\cos 135°}\\ =\frac{\left(\frac{\sqrt{2}}{2}\right)}{\left(-\frac{\sqrt{2}}{2}\right)}\\ =-1\end{array}$

  $\begin{array}{l}\csc 135°\\ =\frac{1}{\sin 135°}\\ =\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}\\ =\sqrt{2}\end{array}$

  $\begin{array}{l}\sec 135°\\ =\frac{1}{\cos 135°}\\ =\frac{1}{\left(-\frac{1}{\sqrt{2}}\right)}\\ =-\sqrt{2}\end{array}$

  $\begin{array}{l}\cot 135°\\ =\frac{1}{\tan 135°}\\ =\frac{1}{-1}\\ =-1\end{array}$