# Prove sinh2x=2sinhx*coshx

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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.**

Here is a video explaining the first step:

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.**

To prove sinh2x=2sinhx*cosh x.

The definitions of hyperbolic functions are as follows:

{e^x+e^-x}/2 = coshx....(1).

(e^x-e^- e^-)}/2 = sinhx. (2).

So we get the product LHS and RHS of the equations :

(e^x+e^-x)*e^x-e^-)/2^2 = coshx*sinhx.

{(e^x)^2 - (e^-x)^2 }/4 = sinhhxcoshx , (a+b)(a-b) = a^2-b62.

(e^2x - e^-2x)/4 = sinhx*coshx, as ((a^m)^n = a^mn.

We multiply by 2:

(e^2x-e^-2x)/2 = 2sinhx*coshx.

sinh2x = 2sinhx*coshx.

Therefore 2sinhx*coshx = sinh2x.