Chapter 7, Problem 88RE

a.

To determine

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

Answer to Problem 88RE

  $\sin \left(\alpha +\beta \right)=\frac{-1-2\sqrt{6}}{6}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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

Calculation:

As it is known that the given identity $\csc \alpha =\frac{1}{\sin \alpha }\text{ and sec}\alpha =\frac{1}{\cos \alpha }$

  $\begin{array}{l}\because \csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)\\ \therefore \sin \alpha =\frac{1}{2}\text{ and }\cos \beta =\frac{-1}{3}\\ and,{\sin }^2\alpha =\frac{1}{4}\text{ and }{\cos }^2\beta =\frac{1}{9}\\ \therefore {\cos }^2\alpha =1-\frac{1}{4}=\frac{3}{4}andsi{n}^2\beta =1-\frac{1}{9}=\frac{8}{9}\\ \therefore \cos \alpha =\pm \frac{\sqrt{3}}{2}\text{ and }\sin \beta =\pm \frac{2\sqrt{2}}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\left(\frac{\pi }{2}<\alpha <\pi \right)\text{ and }\sin \beta =\frac{2\sqrt{2}}{3}\end{array}$ .

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

Therefore,

  $\begin{array}{l}\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{1}{2}\cos \beta =\frac{-1}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\sin \beta =\frac{2\sqrt{2}}{3}\end{array}\right\}\\ =\frac{1}{2}\times \left(\frac{-1}{3}\right)+\frac{2\sqrt{2}}{3}\times -\frac{\sqrt{3}}{2}\\ =-\frac{1}{6}-\frac{2\sqrt{6}}{6}=\frac{-1-2\sqrt{6}}{6}\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha +\beta \right)=\frac{-1-2\sqrt{6}}{6}$ .

b.

To determine

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

Answer to Problem 88RE

  $\cos \left(\alpha +\beta \right)=\frac{\sqrt{3}-\sqrt{2}}{6}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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

Calculation:

As it is known that the given identity $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

Therefore,

  $\begin{array}{l}\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{1}{2}\cos \beta =\frac{-1}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\sin \beta =\frac{2\sqrt{2}}{3}\end{array}\right\}\\ =-\frac{\sqrt{3}}{2}\times \frac{-1}{3}-\frac{1}{2}\times \frac{2\sqrt{2}}{3}\\ =\frac{\sqrt{3}}{6}-\frac{\sqrt{2}}{6}=\frac{\sqrt{3}-\sqrt{2}}{6}\end{array}$

Hence, the given trigonometric function is $\cos \left(\alpha +\beta \right)=\frac{\sqrt{3}-\sqrt{2}}{6}$ .

c.

To determine

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

Answer to Problem 88RE

  $\sin \left(\alpha -\beta \right)=\frac{-1+2\sqrt{6}}{6}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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,

  $\begin{array}{l}\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\sin \alpha =\frac{1}{2}\cos \beta =\frac{-1}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\sin \beta =\frac{2\sqrt{2}}{3}\end{array}\right\}\\ =\frac{1}{2}\times \left(\frac{-1}{3}\right)-\frac{2\sqrt{2}}{3}\times -\frac{\sqrt{3}}{2}\\ =-\frac{1}{6}+\frac{2\sqrt{6}}{6}=\frac{-1+2\sqrt{6}}{6}\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha -\beta \right)=\frac{-1+2\sqrt{6}}{6}$ .

d.

To determine

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

Answer to Problem 88RE

  $\tan \left(\alpha +\beta \right)=\frac{1+2\sqrt{6}}{\sqrt{2}-\sqrt{3}}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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 $\cos \left(\alpha +\beta \right)=\frac{\sqrt{3}-\sqrt{2}}{6}$

And $\sin \left(\alpha +\beta \right)=\frac{-1-2\sqrt{6}}{6}$

Therefore,

  $\begin{array}{l}\tan \left(A+B\right)=\frac{\sin \left(A+B\right)}{\cos \left(A+B\right)}\\ =\frac{\frac{-1-2\sqrt{6}}{6}}{\frac{\sqrt{3}-\sqrt{2}}{6}}=\frac{-1-2\sqrt{6}}{\sqrt{3}-\sqrt{2}}=\frac{1+2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\end{array}$

Hence, the given trigonometric function is $\tan \left(\alpha +\beta \right)=\frac{1+2\sqrt{6}}{\sqrt{2}-\sqrt{3}}$ .

e.

To determine

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

Answer to Problem 88RE

  $\sin \left(2\alpha \right)=-\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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,

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

Hence, the given trigonometric function is $\sin \left(2\alpha \right)=-\frac{\sqrt{3}}{2}$ .

f.

To determine

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

Answer to Problem 88RE

  $\cos \left(2\beta \right)=\frac{-7}{9}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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,

  $\begin{array}{l}\cos \left(2\beta \right)=2{\cos }^2\beta -1\\ \left\{\begin{array}{l}\sin \alpha =\frac{1}{2}\cos \beta =\frac{-1}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\sin \beta =\frac{2\sqrt{2}}{3}\end{array}\right\}\\ \cos \left(2\beta \right)=2{\left(\frac{-1}{3}\right)}^2-1\\ =\frac{2}{9}-1=\frac{-7}{9}\end{array}$

Hence, the given trigonometric function is $\cos \left(2\beta \right)=\frac{-7}{9}$ .

g.

To determine

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

Answer to Problem 88RE

  $\sin \left(\frac{\beta }{2}\right)=\frac{2}{\sqrt{3}}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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,

  $\begin{array}{l}\sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\cos \beta }{2}}\\ \left\{\begin{array}{l}\sin \alpha =\frac{1}{2}\cos \beta =\frac{-1}{3}\\ \cos \alpha =-\frac{\sqrt{3}}{2}\sin \beta =\frac{2\sqrt{2}}{3}\end{array}\right\}\\ \sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\frac{-1}{3}}{2}}=\sqrt{\frac{\frac{3+1}{3}}{2}}=\sqrt{\frac{\frac{4}{3}}{2}}=\sqrt{\frac{4}{6}}=\frac{2}{\sqrt{3}}\end{array}$

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

h.

To determine

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

Answer to Problem 88RE

  $\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{2-\sqrt{3}}{4}}$ .

Explanation of Solution

Given:

The given information is

  $\csc \alpha =2\left(\frac{\pi }{2}<\alpha <\pi \right);\sec \beta =-3\left(\frac{\pi }{2}<\beta <\pi \right)$ .

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,

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

Hence, the given trigonometric function is $\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{2-\sqrt{3}}{4}}$ .