Chapter 4, Problem 30CR

To determine

To find:thesix trigonometric functions at the real number $t=-\frac{\pi }{4}$ if it is possible.

Answer to Problem 30CR

The six trigonometric functions are $\sin \left(-\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}$ , $\cos \left(-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$ , $\tan \left(-\frac{\pi }{4}\right)=-1$ , $\csc \left(-\frac{\pi }{4}\right)=-\sqrt{2}$ , $\sec \left(-\frac{\pi }{4}\right)=\sqrt{2}$ , and $\cot \left(-\frac{\pi }{4}\right)=-1$ .

Explanation of Solution

Definition used:

Let t be a real number and let $\left(x,y\right)$ be the point on the unit circle corresponding to t.

  $\begin{array}{l}\sin t=y\cos t=x\tan t=\frac{y}{x},x\ne 0\\ \csc t=\frac{1}{y},y\ne 0\sec t=\frac{1}{x}\text{,}x\ne 0\cot t=\frac{x}{y},y\ne 0\end{array}$

Calculation:

For every real number t, there exist the corresponding to the point $\left(x,y\right)$ on the unit circle.

Moving clockwise around the unit circle, the value of $t=-\frac{\pi }{4}$ gets in the place of $t=\frac{7\pi }{4}$ .

From Figure 4.13 on page 268, it is observed that the real number $t=\frac{7\pi }{4}$ corresponding to the point $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ .

That is, the real number $t=-\frac{\pi }{4}$ corresponding to the point $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ on the unit circle.

Therefore, the values of $x=\frac{\sqrt{2}}{2}\text{ and}y=-\frac{\sqrt{2}}{2}$ .

The six trigonometric function at the real number is computed as follows,

Substitute $t=-\frac{\pi }{4}$ and $y=-\frac{\sqrt{2}}{2}$ in $\sin t=y$ , the trigonometric function $\sin \left(-\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}$ .

Substitute $t=-\frac{\pi }{4}$ and $x=\frac{\sqrt{2}}{2}$ in $\cos t=x$ , the trigonometric function $\cos \left(-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$ .

Substitute $t=-\frac{\pi }{4}$ , $x=\frac{\sqrt{2}}{2}$ , and $y=-\frac{\sqrt{2}}{2}$ in $\tan t=\frac{y}{x}$ , the trigonometric function is,

  $\begin{array}{c}\tan \left(-\frac{\pi }{4}\right)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\\ =-1\end{array}$

That is, the trigonometric function $\tan \left(-\frac{\pi }{4}\right)=-1$ .

Substitute $t=-\frac{\pi }{4}$ and $y=-\frac{\sqrt{2}}{2}$ in $\csc t=\frac{1}{y}$ , the trigonometric function is,

  $\begin{array}{c}\csc \left(-\frac{\pi }{4}\right)=\frac{1}{-\frac{\sqrt{2}}{2}}\\ =-\frac{2}{\sqrt{2}}\\ =-\sqrt{2}\end{array}$

That is, the trigonometric function $\csc \left(-\frac{\pi }{4}\right)=-\sqrt{2}$ .

Substitute $t=-\frac{\pi }{4}$ and $x=\frac{\sqrt{2}}{2}$ in $\sec t=\frac{1}{x}$ , the trigonometric function is,

  $\begin{array}{c}\sec \left(-\frac{\pi }{4}\right)=\frac{1}{\frac{\sqrt{2}}{2}}\\ =\frac{2}{\sqrt{2}}\\ =\sqrt{2}\end{array}$

That is, the trigonometric function $\sec \left(-\frac{\pi }{4}\right)=\sqrt{2}$ .

Substitute $t=-\frac{\pi }{4}$ , $x=-\frac{\sqrt{2}}{2}$ , and $y=-\frac{\sqrt{2}}{2}$ in $\cot t=\frac{x}{y}$ , the trigonometric function is,

  $\begin{array}{c}\cot \left(-\frac{\pi }{4}\right)=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}\\ =-1\end{array}$

That is, the trigonometric function $\cot \left(-\frac{\pi }{4}\right)=-1$ .

Therefore, the six trigonometric functions are $\sin \left(-\frac{\pi }{4}\right)=-\frac{\sqrt{2}}{2}$ , $\cos \left(-\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$ , $\tan \left(-\frac{\pi }{4}\right)=-1$ , $\csc \left(-\frac{\pi }{4}\right)=-\sqrt{2}$ , $\sec \left(-\frac{\pi }{4}\right)=\sqrt{2}$ , and $\cot \left(-\frac{\pi }{4}\right)=-1$ .