Question

For any x ∈ R, prove that cosh⁴ x - sinh⁴ x = cosh (2x).

Answer 1

We have

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

$LHS=(\frac{e^{x}+e^{-x}}{2})^{2}+(\frac{e^{x}-e^{-x}}{2})^{2}\\\\LHS=\frac{e^{2x}+e^{-2x}+2}{4}+\frac{e^{2x}+e^{-2x}-2}{4}\\\\LHS=\frac{e^{2x}+e^{-2x}+2+e^{2x}+e^{-2x}-2}{4}\\\\LHS=\frac{2e^{2x}+2e^{-2x}}{4}\\\\LHS=\frac{e^{2x}+e^{-2x}}{2}\\\\LHS=cosh(2x)=RHS$

As Required

Answer 2

Answer:

LHS=RHS

Step-by-step explanation:

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