Chapter 13.9, Problem 33WE

(a)

To determine

To find: The value of $\sin \frac{7\pi }{8}$ and $\cos \frac{7\pi }{8}$ .

Answer to Problem 33WE

The values are $\sin \frac{7\pi }{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and $\cos \frac{7\pi }{8}=-\frac{1}{2}\sqrt{2+\sqrt{2}}$ .

Explanation of Solution

Given information:

Use the half angel formula for sine and cosine.

Calculation:

Half-angle formula for the sine:

  $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ .

Half-angle formula for cosine:

  $\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$ .

Thus, $\sin \frac{7\pi }{8}=\sin \frac{\frac{7\pi }{4}}{2}$

Since $\frac{7\pi }{8}$ is a second quadrant angle, then $\sin \frac{7\pi }{8}$ is positive.

Therefore, by applying Half-angle formula for the sine, we get

  $\begin{array}{c}\sin \frac{\frac{7\pi }{4}}{2}=\sqrt{\frac{1-\cos \frac{7\pi }{4}}{2}}\\ =\sqrt{\frac{1-\cos \left(\pi +\frac{3\pi }{4}\right)}{2}}\\ =\sqrt{\frac{1-\left(-\cos \frac{3\pi }{4}\right)}{2}}\\ =\sqrt{\frac{1+\cos \frac{3\pi }{4}}{2}}\\ \end{array}$

  $\begin{array}{c}\sqrt{\frac{1+\cos \frac{3\pi }{4}}{2}}=\sqrt{\frac{1+\cos \left(\pi -\frac{\pi }{4}\right)}{2}}\\ =\sqrt{\frac{1-\cos \frac{\pi }{4}}{2}}\\ =\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}\\ =\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\end{array}$

  $\begin{array}{c}\sqrt{\frac{1+\cos \frac{3\pi }{4}}{2}}=\sqrt{\frac{2-\sqrt{2}}{4}}\\ =\frac{1}{2}\sqrt{2-\sqrt{2}}\end{array}$

Therefore, the value of $\sin \frac{7\pi }{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ .

Calculate the value of $\cos \frac{7\pi }{8}$ .

  $\cos \frac{7\pi }{8}=\cos \frac{\frac{7\pi }{4}}{2}$

Since, $\frac{7\pi }{8}$ is a second quadrant angle, then $\cos \frac{7\pi }{8}$ is negative.

Therefore, by applying Half-angle formula for the cosine, we get

  $\begin{array}{c}\cos \frac{\frac{7\pi }{4}}{2}=-\sqrt{\frac{1+\cos \frac{7\pi }{4}}{2}}\\ =-\sqrt{\frac{1+\cos \left(\pi +\frac{3\pi }{4}\right)}{2}}\\ =-\sqrt{\frac{1-\cos \frac{3\pi }{4}}{2}}\end{array}$

  $\begin{array}{c}-\sqrt{\frac{1-\cos \frac{3\pi }{4}}{2}}=\sqrt{\frac{1-\cos \left(\pi -\frac{\pi }{4}\right)}{2}}\\ =\sqrt{\frac{1+\cos \frac{\pi }{4}}{2}}\\ =-\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\ =-\sqrt{\frac{2+\sqrt{2}}{4}}\\ =-\frac{1}{2}\sqrt{2+\sqrt{2}}\\ \end{array}$

Therefore, the value of $\cos \frac{7\pi }{8}=-\frac{1}{2}\sqrt{2+\sqrt{2}}$ .

(b)

To determine

To find: The value of $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

Answer to Problem 33WE

The value of $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

Explanation of Solution

Given information:

Use answers of (a) $\sin \frac{7\pi }{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and $\cos \frac{7\pi }{8}=-\frac{1}{2}\sqrt{2+\sqrt{2}}$ .

Calculation:

From part (a) we have,

Since,

  $\tan x=\frac{\sin x}{\cos x}$

Thus,

  $\begin{array}{l}\tan \frac{7\pi }{8}=\frac{\sin \frac{7\pi }{8}}{\cos \frac{7\pi }{8}}\\ \tan \frac{7\pi }{8}=\frac{\frac{\cancel{1}}{\cancel{2}}\sqrt{2-\sqrt{2}}}{-\frac{\cancel{1}}{\cancel{2}}\sqrt{2-\sqrt{2}}}\\ \end{array}$

  $=-\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\times \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}}$ Rationalize with $\sqrt{2-\sqrt{2}}$ .

  $\begin{array}{c}\tan \frac{7\pi }{8}=-\sqrt{\frac{{\left(2-\sqrt{2}\right)}^2}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}}\\ \tan \frac{7\pi }{8}=-\sqrt{\frac{\left(4-4\sqrt{2}+2\right)}{2}}\\ =-\sqrt{\frac{\left(6-4\sqrt{2}\right)}{2}}\end{array}$

  $=-\sqrt{\frac{\cancel{2}(3-2\sqrt{2})}{\cancel{2}}}$ Factor out $2$ and cancel it with denominator.

  $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

Therefore, it is showed the value of $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

(c)

To determine

To Find: The value of $\tan \frac{7\pi }{8}=1-\sqrt{2}$ .

Answer to Problem 33WE

Proved the value of $\tan \frac{7\pi }{8}=1-\sqrt{2}$ .

Explanation of Solution

Calculation:

Show that $\tan \frac{7\pi }{8}=1-\sqrt{2}$ by using half-angle formula for tangent.

Half-angle formula for the tangent $\tan \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}$ .

Thus, $\tan \frac{7\pi }{8}=\tan \frac{\frac{7\pi }{4}}{2}$ .

Since $\frac{7\pi }{8}$ is a second quadrant angle, then $\tan \frac{7\pi }{8}$ is negative.

Therefore, by applying Half-angle formula for the tangent, we get

  $\begin{array}{l}\tan \frac{\frac{7\pi }{4}}{2}=-\sqrt{\frac{1-\cos \frac{7\pi }{4}}{1+\cos \frac{7\pi }{4}}}\\ \end{array}$

  $=-\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}$ since $\cos \frac{7\pi }{4}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ .

  $\begin{array}{l}=-\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}\\ \end{array}$

  $=-\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\times \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}}$ Rationalize with $=\sqrt{2-\sqrt{2}}$ .

  $$

  $\begin{array}{l}\tan \frac{7\pi }{8}=-\sqrt{\frac{{(2-\sqrt{2})}^2}{(2+\sqrt{2})(2-\sqrt{2})}}\\ \tan \frac{7\pi }{8}=-\sqrt{\frac{(4-4\sqrt{2}+2)}{2}}\\ =\sqrt{\frac{(6-4\sqrt{2})}{2}}\end{array}$

  $=\sqrt{\frac{\cancel{2}(3-2\sqrt{2})}{\cancel{2}}}$ Factor out 2 and cancel it with denominator.

  $\begin{array}{c}\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}\\ =-\sqrt{{(\sqrt{2})}^2+1^2-2\sqrt{2}}\\ ={\sqrt{-(1-\sqrt{2})}}^2\\ =1-\sqrt{2}\end{array}$

Therefore, it is showed the value of $\tan \frac{7\pi }{8}=1-\sqrt{2}$ .

(d)

To determine

To show: Whether answers of (a) and (b) are equal.

Answer to Problem 33WE

The answers of (a) and (b) are equal.

Explanation of Solution

Calculation:

From part (a), we see that,

  $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

From part (c), we see that,

  $\tan \frac{7\pi }{8}=-\sqrt{3-2\sqrt{2}}.$

Thus, it is proved that the answers to parts (b) and (c) are equal.

Consider the identity,

  $\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\sin x$

Prove the given identity.

According to Pythagorean identity:

  $1+{\tan }^2\frac{x}{2}={\sec }^2\frac{x}{2}$ .

And Double-angle formula for the sine: $\sin 2x=2\sin x\cos x$ .

Consider left hand side,

  $\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

Applying Pythagorean identity,

  $\frac{2\tan \frac{x}{2}}{{\sec }^2\frac{x}{2}}$

Now,

  $\frac{2\tan \frac{x}{2}}{{\sec }^2\frac{x}{2}}=\frac{2\tan \frac{x}{2}}{\left(\frac{1}{{\cos }^2\frac{x}{2}}\right)}$ Since ${\sec }^{2}x=\frac{1}{{\cos }^{2}x}$

  $\begin{array}{l}=2\tan \frac{x}{2}.{\cos }^2\frac{x}{2}\\ =2\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.{\cos }^2\frac{x}{2}\end{array}$

Since $\tan x=\frac{\sin x}{\cos x}$

  $2\sin \frac{x}{2}\cos \frac{x}{2}$

Applying Double-angle formula for the sine we get,

  $\begin{array}{c}2\sin \frac{x}{2}\cos \frac{x}{2}=\sin \cancel{2}\left(\frac{x}{\cancel{2}}\right)\\ =\sin x\\ LHS=RHS\end{array}$

Therefore, the value of (a) and (b) are equal $\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\sin x$ .