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.