Question

Express hyperbolic functionsand in terms of exponential functions.

Short answer

Therefore, the correct answer is $\cos h\sqrt{x}=\frac{e^{\sqrt{x}}+e^{-\sqrt{x}}}{2}$and $\sin h\sqrt{x}=\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2}$.

Step-by-step solution

Determine the formula ofcoshx and  sinhx in exponential functions.

Write the formula of $\cos h\sqrt{x}$in exponential functions.

$\mathbf{cos}\mathit{h}\sqrt{\mathbf{x}}\mathbf{=}\frac{{\mathbf{e}}^{\sqrt{\mathbf{x}}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\sqrt{\mathbf{x}}}}{\mathbf{2}}$……. (1)

Here, $e^{\sqrt{x}}$is a exponential term.

Write the formula of in exponential functions.

$\mathbf{sin}\mathit{h}\sqrt{\mathbf{x}}\mathbf{=}\frac{{\mathbf{e}}^{\sqrt{\mathbf{x}}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\sqrt{\mathbf{x}}}}{\mathbf{2}}$ ……. (2)

Here, $e^{\sqrt{x}}$ is a exponential term.

Determine the difference in coshx and  cosx.

The expansion of$\cos h\sqrt{x}$is given by equation (1).

The expansion of$\sqrt{x}$,

$\cos \sqrt{x}=\frac{e^{i\sqrt{x}}+e^{-i\sqrt{X}}}{2}$

So, observing equations (1) and (2), the difference is of complex term .

Therefore, the correct answer is $\cos h\sqrt{x}=\frac{e^{\sqrt{x}}+e^{-\sqrt{x}}}{2}$and $\sin h\sqrt{x}=\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2}$.