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