# Hyperbolic Functions

### Hyperbolic Functions

Formulas:

i. $$\sinh x=\frac{{{e}^{x}}-{{e}^{-x}}}{2}$$

ii. $$\cosh x=\frac{{{e}^{x}}+{{e}^{-x}}}{2}$$

iii. $$\operatorname{sech}x=\frac{2}{{{e}^{x}}+{{e}^{-x}}}$$

iv. $$\cos echx=\frac{2}{{{e}^{x}}-{{e}^{-x}}}$$

v. $$\tanh x\ \ \ =\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$$

vi. $$\coth x=\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}$$

vii. $${{\cosh }^{2}}x-{{\sinh }^{2}}x=1$$

viii. $$1-{{\tanh }^{2}}x={{\operatorname{sech}}^{2}}x$$

ix. $${{\sinh }^{-1}}x={{\log }_{e}}\{x+\sqrt{{{x}^{2}}+1}\}$$

x. $${{\cosh }^{-1}}x={{\log }_{e}}\{x+\sqrt{{{x}^{2}}-1}\}$$

xi. $${{\tanh }^{-1}}x=\frac{1}{2}{{\log }_{e}}\left( \frac{4x}{1-x} \right)$$

xii. $$\sinh 2x=2\sinh x\cosh x=\frac{2\tanh x}{1-{{\tanh }^{2}}x}$$

xiii. $$\cosh 2x={{\cosh }^{2}}x+{{\sinh }^{2}}x=1=\frac{1+{{\tanh }^{2}}x}{1-{{\tanh }^{2}}x}$$

xiv. $$\tanh (x+y)=\frac{\tanh x+\tanh y}{1-\tanh x.\tanh y}$$

xv. $$\tanh (x-y)=\frac{\tanh x-\tanh y}{1+\tanh x.\tanh y}$$

xvi. $$\coth (x+y)=\frac{\coth x.\coth y+1}{\coth x+\coth y}$$

xvii. $$\coth (x-y)=\frac{\coth x.\coth y-1}{\coth x-\coth y}$$

Example: If $$\mu ={{\log }_{e}}\left\{ \tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right\}$$ the prove that cosh μ = secθ.

Solution: $$\mu ={{\log }_{e}}\left\{ \tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right\}$$,

cosh μ = secθ

$$\mu ={{\log }_{e}}\left\{ \tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right\}$$,

$${{e}^{\mu }}=\tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right)$$,

$${{e}^{-\mu }}=\cot \left( \frac{\pi }{4}+\frac{\theta }{2} \right)$$,

$$\cosh \mu =\frac{{{e}^{\mu }}+{{e}^{-\mu }}}{2}=\frac{1}{2}\left( \tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right)+\cot \left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right)$$,

$$=\frac{1}{2}\left( \frac{\sin \left( \frac{\pi }{4}+\frac{\theta }{2} \right)}{\cos \left( \frac{\pi }{4}+\frac{\theta }{2} \right)}+\frac{\cos \left( \frac{\pi }{4}+\frac{\theta }{2} \right)}{\sin \left( \frac{\pi }{4}+\frac{\theta }{2} \right)} \right)$$,

$$=\frac{1}{2}\left( \frac{{{\sin }^{2}}\left( \frac{\pi }{4}+\frac{\theta }{2} \right)+{{\cos }^{2}}\left( \frac{\pi }{4}+\frac{\theta }{2} \right)}{\sin \left( \frac{\pi }{4}+\frac{\theta }{2} \right)\cos \left( \frac{\pi }{4}+\frac{\theta }{2} \right)} \right)$$,

$$=\frac{1}{2}\left( \frac{1}{\sin \left( \frac{\pi }{4}+\frac{\theta }{2} \right)\cos \left( \frac{\pi }{4}+\frac{\theta }{2} \right)} \right)$$,

$$=\frac{1}{2\sin \left( \frac{\pi }{4}+\frac{\theta }{2} \right)\cos \left( \frac{\pi }{4}+\frac{\theta }{2} \right)}$$,

$$=\frac{1}{\sin \left\{ 2\left( \frac{\pi }{4}+\frac{\theta }{2} \right) \right\}}$$,

$$=\frac{1}{\sin \left( \frac{\pi }{2}+\theta \right)}$$,

$$=\frac{1}{\cos \theta }$$,

= secθ

Hence proved cosh μ = secθ.