Chapter A6, Problem 11E

(a)

To determine

To show: the given trigonometric function..

Explanation of Solution

Given:

  $\sinh \left(x+y\right)=\sinh x\text{ cosh y+cosh }x\text{ sinh y}\text{.}$

Proof:

to prove,

  $\begin{array}{l}\text{sinh }2x=2\sinh x\cosh x.\\ \Rightarrow \text{sinh }2x=\sinh \left(x+x\right).\\ \Rightarrow \text{sinh }2x=\sinh x\cosh x+\text{cosh }x\text{ sinh }x\left(\text{using}\left(1\right)\right).\\ \Rightarrow \text{sinh }2x=2\sinh x\cosh x.\end{array}$

Hence proved .

Conclusion:

Therefore, $\text{sinh }2x=2\sinh x\cosh x.$ is proved.

(b)

To show: the given trigonometric function.

Given: $\cosh \left(x+y\right)=\cosh x\text{ cosh y+sinh }x\text{ sinh y}\text{.}$

Proof:

  $\begin{array}{l}\text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x.\\ \Rightarrow \text{cosh 2}x=\text{cosh}\left(x+x\right).\\ \Rightarrow \text{cosh 2}x=\cosh x\cosh x+\sinh x\sinh x\left(\text{using}\left(2\right)\right).\\ \Rightarrow \text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x.\end{array}$

Conclusion:

Therefore, $\text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x$ is proved.

(b)

To determine

To show: the given trigonometric function.

Explanation of Solution

Given: $\cosh \left(x+y\right)=\cosh x\text{ cosh y+sinh }x\text{ sinh y}\text{.}$

Proof:

  $\begin{array}{l}\text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x.\\ \Rightarrow \text{cosh 2}x=\text{cosh}\left(x+x\right).\\ \Rightarrow \text{cosh 2}x=\cosh x\cosh x+\sinh x\sinh x\left(\text{using}\left(2\right)\right).\\ \Rightarrow \text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x.\end{array}$

Conclusion:

Therefore, $\text{cosh 2}x={\text{cosh}}^{2}x+{\sinh }^{2}x$ is proved.