Question

Prove that
sinh2x=2sinh^x-1​

Answer 1

Hyperbolic functions

$sinhx = \dfrac{e ^{x} - e ^{ - x} }{2} \\ \\ coshx = \dfrac{e ^{x} + e ^{ - x} }{2}$

To prove : cosh(2x) =2sinh^2x - 1

Now,

LHS

$cos h(2x) = \dfrac{e ^{2x} + e ^{ - 2x} }{2}$

RHS

$2sinh ^{2} x - 1 \\ \\ = 2( \dfrac{ {e}^{x } + {e}^{ - x} }{2} ) ^{2} - 1 \\ \\ = 2( \dfrac{e ^{2x} + {e}^{ - 2x } +2 }{4} ) - 1 \\ \\ = ( \dfrac{e ^{2x} + {e}^{ - 2x } +2 }{2} ) - 1 \\ \\ = ( \dfrac{e ^{2x} + {e}^{ - 2x } +2 - 2}{2} ) \\ \\ = \frac{ {e}^{2x} + {e}^{ - 2x} }{2} = cosh(2x)$

Hence proved that, cosh(2x) =2sinh^2x - 1