# Thread: Trig Identities

1. ## Trig Identities

Anyone give me a leg up on verifying the following identity?

sinh(x+y) = sinh x cosh y + cosh x sinh y

2. Hello, kaiser0792!

Use the definitions: . $\begin{array}{ccc}\sinh z &=& \dfrac{e^z - e^{-z}}{2} \\ \\[-3mm]\cosh z &=& \dfrac{e^z + e^{-z}}{2} \end{array}$

Verify: . $\sinh(x+y) \:=\: \sinh x \cosh y + \cosh x \sinh y$

The right side is:

$\sinh x\cosh y + \cosh x \sinh y$

. . . . $=\;\frac{e^x-e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2} + \frac{e^x + e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}$

. . . . $=\;\frac{e^{x+y} + e^{x-y} - e^{y-x} - e^{-x-y}}{4} + \frac{e^{x+y} - e^{x-y} + e^{y-x} - e^{-x-y}}{4}$

. . . . $=\; \frac{2e^{x+y} - 2e^{-(x+y)}}{4}$

. . . . $=\;\frac{e^{x+y} - e^{-(x+y)}}{2}$

. . . . $=\;\sinh(x+y)$

3. ## Thanks Soroban

Excellent Soroban, I didn't think of it.
Thank you