Chapter 6, Problem 11CT

To determine

The exact value of the expression, $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ .

Answer to Problem 11CT

Solution:

The exact value of the expression $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ is $2$ .

Explanation of Solution

Given information:

The expression, $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$

Let, the expression be $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ .

To determine exact value of the expression, $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ , first find out the exact value of $\sin \left(\frac{\pi }{2}\right)$ and $\tan \left(\frac{19\pi }{4}\right)$

Write $\frac{19\pi }{4}$ as $\frac{19\pi }{4}=4\pi +\frac{3\pi }{4}=2\pi +2\pi +\frac{3\pi }{4}$ .

According to periodic property, $\tan \left(\frac{19\pi }{4}\right)=\tan \left(2\pi +\left(2\pi +\frac{3\pi }{4}\right)\right)=\tan \left(2\pi +\frac{3\pi }{4}\right)=\tan \left(\frac{3\pi }{4}\right)$ .

To determine the exact value of the expression, $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ , first find out the exact value of $\sin \left(\frac{\pi }{2}\right)$ and $\tan \left(\frac{3\pi }{4}\right)$ using unit circle.

The point $P$ on the unit circle that corresponds to $\theta =\frac{\pi }{2}$ is $P=\left(0,1\right)$ ,

So $\sin \frac{\pi }{2}=y=1$ .

The point $P$ on the unit circle that corresponds to $\theta =\frac{3\pi }{4}$ is $P=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ , so $\sin \frac{3\pi }{4}=y=\frac{\sqrt{2}}{2}$ .

  $\cos \frac{3\pi }{4}=x=-\frac{\sqrt{2}}{2}$

Then, $\tan \left(\frac{3\pi }{4}\right)=\frac{\sin \left(\frac{3\pi }{4}\right)}{\cos \left(\frac{3\pi }{4}\right)}=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=-1$

Thus, $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}=1-(-1)=1+1=2$ .

Thus, the exact value of the expression $\sin \frac{\pi }{2}-\tan \frac{19\pi }{4}$ is $2$ .