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


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

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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Hint: $2x=x+x$. – Hans Lundmark Oct 28 '12 at 21:34
where did the hyperbolic cosign go? – dsta Oct 28 '12 at 21:34

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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I don't understand, how does this apply to my problem? – dsta Oct 28 '12 at 21:43
I need to figure out this problem without using exp. I'm basically filling in the ? in WebAssign. – dsta Oct 28 '12 at 21:45
ok so $\sinh 2x = \sinh (x + x)$ and $\sinh x \cosh x + \cosh x \sinh x$, what happens in the end? – dsta Oct 28 '12 at 21:54
@dsta Yes. You can then expand to get what you want. – user17762 Oct 28 '12 at 21:55
@dsta: You need to start getting used to accept answers. – ILoveMath Oct 28 '12 at 22:52

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