Chapter 13, Problem 12CT

(a)

To determine

To find: The value of expression $\sin 2\alpha$ .

Answer to Problem 12CT

The value of the expression is $\sin 2\alpha =-\frac{240}{289}$ .

Explanation of Solution

Given information:

The given expression is $\sin \alpha =\frac{8}{17}$ and $\cos \alpha <0$ .

Calculation:

Consider the function.

  $\sin \alpha =\frac{8}{17}$ and $\cos \alpha <0$ .

By apply Pythagorean theorem,

  $\begin{array}{c}{\sin }^2\alpha +{\cos }^2\alpha =1\\ \cos \alpha =\pm \sqrt{1-{\sin }^2\alpha }\\ =\pm \sqrt{1-{\left(\frac{8}{17}\right)}^2}\\ =\pm \frac{15}{17}\end{array}$

Now,

  $\begin{array}{c}\sin 2\alpha =2\sin \alpha \cos \alpha \\ =2\left(\frac{8}{17}\right)·\left(\frac{-15}{17}\right)\\ =-\frac{240}{289}\end{array}$

Therefore, the value of the expression is $\sin 2\alpha =-\frac{240}{289}$ .

(b)

To determine

To find: The value of expression $\cos 2\alpha$ .

Answer to Problem 12CT

The value of the expression is $\cos 2\alpha =-\frac{161}{289}$ .

Explanation of Solution

Given information:

The given expression is $\sin \alpha =\frac{8}{17}$ and $\cos \alpha <0$ .

Calculation:

Consider the function.

  $\sin \alpha =\frac{8}{17}$ and $\cos \alpha <0$ .

By apply Pythagorean theorem,

  $\begin{array}{c}\sin 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha \\ ={\left(-\frac{15}{17}\right)}^2-{\left(\frac{8}{17}\right)}^2\\ =-\frac{161}{289}\end{array}$

Therefore, the value of the expression is $\cos 2\alpha =-\frac{161}{289}$ .

(c)

To determine

To find: The value of expression $\sin \frac{\alpha }{2}$ .

Answer to Problem 12CT

The value of the expression is $\sin \frac{\alpha }{2}=\frac{4\sqrt{17}}{17}$ .

Explanation of Solution

Given information:

The given expression is $\sin \alpha =\frac{8}{17}$ and $\cos \alpha <0$ .

Calculation:

Consider the function.

  $\begin{array}{c}\sin \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}}\\ =\sqrt{\frac{16}{17}}\\ =\frac{4\sqrt{17}}{17}\end{array}$

Therefore, the value of the expression is $\sin \frac{\alpha }{2}=\frac{4\sqrt{17}}{17}$ .