Question

Prove the identity\({(\cosh x + \sinh x)^n} = \cosh nx + \sinh nx\) where n is any real number.

Short answer

The identity \({(\cosh x + \sinh x)^n} = \cosh nx + \sinh nx\) is proved.

Step-by-step solution

Given

The identity is \({(\cosh x + \sinh x)^n} = \cosh nx + \sinh nx\).

The concept of the hyperbolic function

(1) The hyperbolic sine function:\(\user1{sinhx = }\frac{{{\user1{e}^\user1{x}}\user1{ - }{\user1{e}^{\user1{ - x}}}}}{\user1{2}}\)

(2) The hyperbolic cosine function:\(\user1{coshx = }\frac{{{\user1{e}^\user1{x}}\user1{ + }{\user1{e}^{\user1{ - x}}}}}{\user1{2}}\)

(3) The hyperbolic tangent function:\(\user1{tanhx = }\frac{{\user1{sinhx}}}{{\user1{coshx}}}\)

Use the formulas and prove

Consider, \({(\cosh x + \sinh x)^n}\).

Apply the definitions mentioned above and simplify the terms as shown below.

\(\begin{array}{c}{(\cosh x + \sinh x)^n} = {\left( {\frac{{{e^x} + {e^{ - x}}}}{2} + \frac{{{e^x} - {e^{ - x}}}}{2}} \right)^n}\\ = {\left( {\frac{{{e^x} + {e^{ - x}} + {e^x} - {e^{ - x}}}}{2}} \right)^n}\\ = {\left( {\frac{{2{e^x}}}{2}} \right)^n}\\ = {e^{nx}}\end{array}\)

Add and subtract the value \({e^{nx}}\) on numerator.

\(\begin{array}{c}{(\cosh x + \sinh x)^n} = {e^{nx}}\\ = \frac{{{e^{nx}} + {e^{ - nx}} + {e^{nx}} - {e^{ - nx}}}}{2}\\ = \frac{{{e^{nx}} + {e^{ - nx}}}}{2} + \frac{{{e^{nx}} - {e^{ - nx}}}}{2}\\ = \cosh nx + \sinh nx\end{array}\)

Hence, the identity \({(\cosh x + \sinh x)^n} = \cosh nx + \sinh nx\) is proved.