Question

How do you prove $\cosh (2x) = {\cosh ^2}x + {\sinh ^2}x$ ?

Answer 1

Hint : First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the following formula
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} \\\ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} \;$
to evaluate and then we will further simplify this expression form and hence evaluate the value of the term.

Complete step-by-step answer :
We will start off by using the formula
$\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} \\\ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} \\\$ .
Here, we will start by evaluating the right-hand side of the equation.
Hence, the equation will become,

$$= {\cosh ^2}x + {\sinh ^2}x \\\ = {\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)^2} + {\left( {\dfrac{{{e^x} - {e^{ - x}}}}{2}} \right)^2} \\\ = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} + 2}}{4}} \right) + \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} - 2}}{4}} \right) \\\ = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}} + 2 + {e^{2x}} + {e^{ - 2x}} - 2}}{4}} \right) \\\ = \left( {\dfrac{{2{e^{2x}} + 2{e^{ - 2x}}}}{4}} \right) \\\ = 2\left( {\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{4}} \right) \\\ = \left( {\dfrac{{{e^{2x}} + {e^{ - 2x}}}}{2}} \right) \\\ = \cosh (2x) \;$$

Hence, LHS = $\cosh (2x)$
And we know that RHS = $\cosh (2x)$
Therefore, RHS=LHS
Hence, proved.

Note : n mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. Just as the points form a circle with a unit radius, the points form the right half of the unit hyperbola