Chapter 7, Problem 90RE

a.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\sin \left(\alpha +\beta \right)=-1$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-2,\frac{\pi }{2}<\alpha <\pi ;\cot \beta =-2,\frac{\pi }{2}<\beta <\pi$ .

And the given trigonometric function is $\sin \left(\alpha +\beta \right)$ .

Calculation:

As it is known that the given identity $tan\alpha =\frac{\sin \alpha }{\cos \alpha }$

  $\begin{array}{l}\tan \alpha =-2,\left(\frac{\pi }{2}<\alpha <\pi \right);\cot \beta =-2,\frac{\pi }{2}<\beta <\pi \\ \because \tan \alpha =-2,\left(\frac{\pi }{2}<\alpha <\pi \right)\\ \therefore \sin \alpha =\frac{1}{\sqrt{5}}\text{ and }\cos \alpha =\frac{-2}{\sqrt{5}}\\ \cot \beta =-2\left(\pi <\beta <\frac{3\pi }{2}\right)\\ \therefore \sin \beta =\frac{1}{\sqrt{5}}\text{ and }\cos \beta =\frac{-2}{\sqrt{5}}\end{array}$ .

As it is known that the given identity $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$ .

Therefore,

  $\begin{array}{l}\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ =\frac{2}{\sqrt{5}}\times \frac{-2}{\sqrt{5}}+-\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{5}}\\ =\frac{-4}{5}+\frac{-1}{5}=\frac{-5}{5}=-1\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha +\beta \right)=-1$ .

b.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\cos \left(\alpha +\beta \right)=0$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\cos \left(\alpha +\beta \right)$ .

Calculation:

As it is known that the given identity $tan\alpha =\frac{\sin \alpha }{\cos \alpha }$

Therefore,

  $\begin{array}{l}\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ =-\frac{1}{\sqrt{5}}\times \frac{-2}{\sqrt{5}}-\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}\\ =\frac{-2}{5}-\frac{-2}{5}=0\end{array}$

Hence, the given trigonometric function is $\cos \left(\alpha +\beta \right)=0$ .

c.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\sin \left(\alpha -\beta \right)=\frac{-3}{5}$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\sin \left(\alpha -\beta \right)$ .

Calculation:

As it is known that the given identity $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$ .

Therefore,

  $\begin{array}{l}\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ =\frac{2}{\sqrt{5}}\times \frac{-2}{\sqrt{5}}-\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{5}}\\ =\frac{-4}{5}-\frac{-1}{5}=\frac{-3}{5}\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha -\beta \right)=\frac{-3}{5}$ .

d.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\tan \left(\alpha +\beta \right)=\infty$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\tan \left(\alpha +\beta \right)$ .

Calculation:

As it is known that the given identity $\tan \left(A+B\right)=\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)}$ .

As $\cos \left(\alpha +\beta \right)=0$

And $\sin \left(\alpha +\beta \right)=-1$

Therefore,

  $\begin{array}{l}\tan \left(A+B\right)=\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)}\\ =\frac{-1}{0}=\infty \end{array}$

Hence, the given trigonometric function is $\tan \left(\alpha +\beta \right)=\infty$ .

e.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\sin \left(2\alpha \right)=\frac{-4}{5}$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\sin \left(2\alpha \right)$ .

Calculation:

As it is known that the given identity $\sin \left(2A\right)=2\sin A\cos A$ .

Therefore,

  $\begin{array}{l}\sin \left(2A\right)=2\sin \alpha \cos \alpha \\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ \sin \left(2A\right)=2\sin \alpha \cos \alpha \\ =2\times \frac{2}{\sqrt{5}}\times \frac{-1}{\sqrt{5}}=\frac{-4}{5}\end{array}$

Hence, the given trigonometric function is $\sin \left(2\alpha \right)=\frac{-4}{5}$ .

f.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\cos \left(2\beta \right)=\frac{3}{5}$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\cos \left(2\beta \right)$ .

Calculation:

As it is known that the given identity $\cos \left(2A\right)={\cos }^{2}A-{\sin }^{2}A$ .

Therefore,

  $\begin{array}{l}\cos \left(2\beta \right)=2{\cos }^2\beta -1\\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ \cos \left(2\beta \right)=2{\left(\frac{-2}{\sqrt{5}}\right)}^2-1\\ =\frac{8}{5}-1=\frac{3}{5}\end{array}$

Hence, the given trigonometric function is $\cos \left(2\beta \right)=\frac{3}{5}$ .

g.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\sin \left(\frac{\beta }{2}\right)=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}}$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\sin \left(\frac{\beta }{2}\right)$ .

Calculation:

As it is known that the given identity $\sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\cos \beta }{2}}$ .

Therefore,

  $\begin{array}{l}\sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\cos \beta }{2}}\\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ \sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\frac{-2}{\sqrt{5}}}{2}}=\sqrt{\frac{\frac{\sqrt{5}+2}{\sqrt{5}}}{2}}=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}}\left\{\text{sin}\left(\frac{\beta }{2}\right)\text{ is positive}\right\}\end{array}$

Hence, the given trigonometric function is $\sin \left(\frac{\beta }{2}\right)=\sqrt{\frac{\sqrt{5}+2}{2\sqrt{5}}}$ .

h.

To determine

To find: the exact value of the given trigonometric function by using the given information.

Answer to Problem 90RE

  $\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\sqrt{5}}{2\sqrt{5}}}$ .

Explanation of Solution

Given:

The given information is

  $\tan \alpha =-\frac{4}{3},\frac{\pi }{2}<\alpha <\pi ;\cot \beta =\frac{12}{5},\pi <\beta <\frac{3\pi }{2}$ .

And the given trigonometric function is $\cos \left(\frac{\alpha }{2}\right)$ .

Calculation:

As it is known that the given identity $\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\cos \alpha }{2}}$ .

Therefore,

  $\begin{array}{l}\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\cos \alpha }{2}}\\ \left\{\begin{array}{l}\sin \alpha =\frac{2}{\sqrt{5}},\cos \beta =\frac{-2}{\sqrt{5}}\\ \cos \alpha =-\frac{1}{\sqrt{5}},sin\beta =\frac{1}{\sqrt{5}}\end{array}\right\}\\ \cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\frac{1}{\sqrt{5}}}{2}}=\sqrt{\frac{1+\sqrt{5}}{2\sqrt{5}}}\left\{\text{since }\cos \left(\frac{\alpha }{2}\right)\text{is postive}\right\}\end{array}$

Hence, the given trigonometric function is $\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\sqrt{5}}{2\sqrt{5}}}$ .