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:\({sinhx = }\frac{{{{e}^{x}}{ - }{{e}^{{ - x}}}}}{{2}}\)

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

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

Use the formulas and prove

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

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

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

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

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

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