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(Note: The ? is where I have to fill in something)

Prove:

$ \begin{align} \sinh2x & = 2\sinh x\cosh x\\ \sinh2x & = \sinh (x + ?)\\ & = \sinh x(?) + \cosh x \sinh x\\ & =\space?\\ \end{align} $

I know that there is a way of proving it by using the basic definitions of $\cosh$ and $\sinh$, but that is not what the question is asking.

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    where did the hyperbolic cosign go?2012-10-28

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HINT: Recall that $\cosh(y) = \dfrac{\exp(y) + \exp(-y)}2$ and $\sinh(y) = \dfrac{\exp(y) - \exp(-y)}2$ Make us of the identity $(a^2 - b^2) = (a+b)(a-b)$ Hence, $\sinh(2x) = \dfrac{\exp(2x) - \exp(-2x)}2 = \dfrac12 \left((\exp(x))^2 - (\exp(-x))^2\right)$ Now you should be able to finish it off.

EDIT $\sinh(a+b) = \sinh(a) \cos(b) + \sinh(b) \cosh(a)$ This is similar to the formula $\sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a)$ In your problem, take $a=b=x$.

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    @dsta Just add the two terms. Since both terms are the same terms you get $2 \sinh(x) \cosh(x)$2012-10-28