Prove sinh2x = 2sinhx*coshx .

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We'll write the formula for each term of the given identity:

sinh2x = (1/2)(e^2x - e^-2x)

sinh x = (1/2)(e^x - e^-x)

cosh x = (1/2)(e^x + e^-x)

We'll re-write the identity that has to be demonstrated:

(1/2)(e^2x - e^-2x) = 2*(1/2)*(1/2)(e^x - e^-x)*(e^x + e^-x)

We'll simplify and we'll get:

(1/2)(e^2x - e^-2x) = (1/2)(e^x - e^-x)*(e^x + e^-x)

We'll re-write the product from the right side as a difference of squares:

(1/2)(e^2x - e^-2x) = (1/2)[(e^x)^2 - (e^-x)^2]

We'll multiply the exponents and we'll get:

**(1/2)(e^2x - e^-2x) = (1/2)(e^2x - e^-2x) q.e.d.**

You need to write `sinh 2x` as it follows:

`sinh 2x = sinh (x + x)`

Using the following trigonometric identity yields:

`sinh(alpha + beta) = sinh alpha*cosh beta + cosh alpha*sinh beta`

Reasoning by analogy yields:

`sinh (x + x) = sinh x*cosh x + cosh x*sinh x`

`sinh (x + x) = 2 sinh x*cosh x`

**Hence, testing the identity `sinh 2x = 2 sinh x*cosh x` yields that it holds.**

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