Chapter 7, Problem 82RE

a.

To determine

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

Answer to Problem 82RE

  $\sin \left(\alpha +\beta \right)=\frac{-33}{65}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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

Calculation:

As it is known that the given identity ${\sin }^2\alpha +{\cos }^2\alpha =1$

  $\begin{array}{l}\Rightarrow \left(\because \cos \alpha =\frac{4}{5}\Rightarrow {\cos }^2\alpha =\frac{16}{25}\right)\\ \Rightarrow \frac{16}{25}+{\sin }^2\alpha =1\\ \therefore {\sin }^2\alpha =1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{25}\\ sin\alpha =\sqrt{\frac{9}{25}}=\frac{3}{5}\end{array}$ .

  $\begin{array}{l}\Rightarrow \left(\because \cos \beta =\frac{5}{13}\Rightarrow {\cos }^2\beta =\frac{25}{169}\right)\\ \Rightarrow {\sin }^2\beta +\frac{25}{169}=1\\ \therefore {\sin }^2\beta =1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}\\ \sin \beta =\sqrt{\frac{144}{169}}=\pm \frac{12}{13}\end{array}$

As $\sin \beta$ is in 3rd quadrant, therefore $\sin \beta =\frac{-12}{13}$

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

Therefore,

  $\begin{array}{l}\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\because \cos \alpha =\frac{4}{5},cos\beta =\frac{5}{13}\left\{\text{given}\right\}\\ \therefore \sin \alpha =\frac{3}{5},sin\beta =\frac{-12}{13}\end{array}\right\}\\ =\frac{3}{5}\times \frac{5}{13}+\frac{4}{5}\times \frac{-12}{13}\\ =\frac{15}{65}+\frac{-48}{65}=\frac{-33}{65}\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha +\beta \right)=\frac{-33}{65}$ .

b.

To determine

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

Answer to Problem 82RE

  $\cos \left(\alpha +\beta \right)=\frac{56}{65}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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

Calculation:

As it is known that the given identity ${\sin }^2\alpha +{\cos }^2\alpha =1$

  $\begin{array}{l}\Rightarrow \left(\because \cos \alpha =\frac{4}{5}\Rightarrow {\cos }^2\alpha =\frac{16}{25}\right)\\ \Rightarrow \frac{16}{25}+{\sin }^2\alpha =1\\ \therefore {\sin }^2\alpha =1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{25}\\ sin\alpha =\sqrt{\frac{9}{25}}=\frac{3}{5}\end{array}$ .

  $\begin{array}{l}\Rightarrow \left(\because \cos \beta =\frac{5}{13}\Rightarrow {\cos }^2\beta =\frac{25}{169}\right)\\ \Rightarrow {\sin }^2\beta +\frac{25}{169}=1\\ \therefore {\sin }^2\beta =1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}\\ \sin \beta =\sqrt{\frac{144}{169}}=\pm \frac{12}{13}\end{array}$

As $\sin \beta$ is in 3rd quadrant, therefore $\sin \beta =\frac{-12}{13}$

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

Therefore,

  $\begin{array}{l}\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \left\{\begin{array}{l}\because \sin \alpha =\frac{4}{5},sin\beta =\frac{5}{13}\left\{\text{given}\right\}\\ \therefore \cos \alpha =\frac{3}{5},\cos \beta =\frac{-12}{13}\end{array}\right\}\\ =-\frac{4}{5}\times \frac{5}{13}-\frac{3}{5}\times \frac{-12}{13}\\ =\frac{20}{65}+\frac{36}{65}+=\frac{56}{65}\end{array}$

Hence, the given trigonometric function is $\cos \left(\alpha +\beta \right)=\frac{56}{65}$ .

c.

To determine

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

Answer to Problem 82RE

  $\sin \left(\alpha -\beta \right)=\frac{63}{65}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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

Calculation:

As it is known that the given identity ${\sin }^2\alpha +{\cos }^2\alpha =1$

  $\begin{array}{l}\Rightarrow \left(\because \cos \alpha =\frac{4}{5}\Rightarrow {\cos }^2\alpha =\frac{16}{25}\right)\\ \Rightarrow \frac{16}{25}+{\sin }^2\alpha =1\\ \therefore {\sin }^2\alpha =1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{25}\\ sin\alpha =\sqrt{\frac{9}{25}}=\frac{3}{5}\end{array}$ .

  $\begin{array}{l}\Rightarrow \left(\because \cos \beta =\frac{5}{13}\Rightarrow {\cos }^2\beta =\frac{25}{169}\right)\\ \Rightarrow {\sin }^2\beta +\frac{25}{169}=1\\ \therefore {\sin }^2\beta =1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}\\ \sin \beta =\sqrt{\frac{144}{169}}=\pm \frac{12}{13}\end{array}$

As $\sin \beta$ is in 3rd quadrant, therefore $\sin \beta =\frac{-12}{13}$

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

Therefore,

  $\begin{array}{l}\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ \left\{\begin{array}{l}\because \cos \alpha =\frac{4}{5},cos\beta =\frac{5}{13}\left\{\text{given}\right\}\\ \therefore \sin \alpha =\frac{3}{5},sin\beta =\frac{-12}{13}\end{array}\right\}\\ =\frac{3}{5}\times \frac{5}{13}-\frac{4}{5}\times \frac{-12}{13}\\ =\frac{15}{65}-\frac{-48}{65}=\frac{63}{65}\end{array}$

Hence, the given trigonometric function is $\sin \left(\alpha -\beta \right)=\frac{63}{65}$ .

d.

To determine

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

Answer to Problem 82RE

  $\tan \left(\alpha +\beta \right)=-\frac{33}{56}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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

Calculation:

As it is known that the given identity ${\sin }^2\alpha +{\cos }^2\alpha =1$

  $\begin{array}{l}\Rightarrow \left(\because \cos \alpha =\frac{4}{5}\Rightarrow {\cos }^2\alpha =\frac{16}{25}\right)\\ \Rightarrow \frac{16}{25}+{\sin }^2\alpha =1\\ \therefore {\sin }^2\alpha =1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{25}\\ sin\alpha =\sqrt{\frac{9}{25}}=\frac{3}{5}\end{array}$ .

  $\begin{array}{l}\Rightarrow \left(\because \cos \beta =\frac{5}{13}\Rightarrow {\cos }^2\beta =\frac{25}{169}\right)\\ \Rightarrow {\sin }^2\beta +\frac{25}{169}=1\\ \therefore {\sin }^2\beta =1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}\\ \sin \beta =\sqrt{\frac{144}{169}}=\pm \frac{12}{13}\end{array}$

As $\sin \beta$ is in 3rd quadrant, therefore $\sin \beta =\frac{-12}{13}$

As it is known that the given identity $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ .

Therefore,

  $\begin{array}{l}\tan \left(A+B\right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ \left\{\begin{array}{l}\because \cos \alpha =\frac{4}{5},cos\beta =\frac{5}{13}\left\{\text{given}\right\}\\ \therefore \sin \alpha =\frac{3}{5},sin\beta =\frac{-12}{13}\end{array}\right\}\end{array}$

  $\begin{array}{l}\tan \left(\alpha +\beta \right)=\frac{\frac{3}{4}+\frac{-12}{5}}{1-\left(\frac{3}{4}\times \frac{-12}{5}\right)}\\ =\frac{\frac{15-48}{20}}{1-\left(\frac{-36}{20}\right)}\\ =\frac{-\frac{33}{20}}{\frac{56}{36}}=-\frac{33}{56}\end{array}$

Hence, the given trigonometric function is $\tan \left(\alpha +\beta \right)=-\frac{33}{56}$ .

e.

To determine

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

Answer to Problem 82RE

  $\sin \left(2\alpha \right)=\frac{24}{25}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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,

  $\begin{array}{l}\sin \left(2A\right)=2\sin \alpha \cos \alpha \\ \left(\because \cos \alpha =\frac{4}{5}\Rightarrow \sin \alpha =\frac{3}{5}\right)\\ \sin \left(2A\right)=2\sin \alpha \cos \alpha \\ =2\times \frac{4}{5}\times \frac{3}{5}=\frac{24}{25}\end{array}$

Hence, the given trigonometric function is $\sin \left(2\alpha \right)=\frac{24}{25}$ .

f.

To determine

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

Answer to Problem 82RE

  $\cos \left(2\beta \right)=-\frac{119}{169}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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

Calculation:

As it is known that the given identity $\cos \left(2A\right)=2{\cos }^{2}A-1$ .

Therefore,

  $\begin{array}{l}\cos \left(2\beta \right)=2{\cos }^2\beta -1\\ \left\{\because \cos \beta =\frac{5}{13}\right\}\\ \cos \left(2\beta \right)=2{\left(\frac{5}{13}\right)}^2-1\\ =\frac{50}{169}-1=-\frac{119}{169}\end{array}$

Hence, the given trigonometric function is $\cos \left(2\beta \right)=-\frac{119}{169}$ .

g.

To determine

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

Answer to Problem 82RE

  $\sin \left(\frac{\beta }{2}\right)=-\frac{2\sqrt{2}}{\sqrt{13}}$ .

Explanation of Solution

Given:

The given information is

  $\cos \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\cos \beta =\frac{5}{13},-\frac{\pi }{2}<\beta <0$ .

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,

  $\begin{array}{l}\sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\cos \beta }{2}}\\ \left\{\cos \beta =\frac{5}{13}(given)\right\}\\ \sin \left(\frac{\beta }{2}\right)=\pm \sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13-5}{13}}{2}}=\sqrt{\frac{\frac{8}{13}}{2}}=-\frac{2\sqrt{2}}{\sqrt{13}}\end{array}$

Hence, the given trigonometric function is $\sin \left(\frac{\beta }{2}\right)=-\frac{2\sqrt{2}}{\sqrt{13}}$ .

h.

To determine

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

Answer to Problem 82RE

  $\cos \left(\frac{\alpha }{2}\right)=\frac{1}{2\sqrt{13}}$ .

Explanation of Solution

Given:

The given information is

  $\sin \alpha =\frac{4}{5},0<\alpha <\frac{\pi }{2};\sin \beta =\frac{5}{13},\frac{\pi }{2}<\beta <\pi$ .

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

Calculation:

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

Therefore,

  $\begin{array}{l}\cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\cos \alpha }{2}}\\ \left\{\because \cos \alpha =\frac{4}{5},\right\}\\ \cos \left(\frac{\alpha }{2}\right)=\sqrt{\frac{1+\frac{4}{5}}{2}}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}}\end{array}$

Hence, the given trigonometric function is $\cos \left(\frac{\alpha }{2}\right)=\frac{3}{\sqrt{10}}$ .