## anonymous 5 years ago Prove cosh(2x) = cosh^(2)x + sin^(2)x

1. anonymous

its just going back to the definitions

2. anonymous

$\cosh (nx) = \frac{1}{2} ( e^{nx} + e^{-nx} )$

3. anonymous

$\sinh(nx) = \frac{1}{2}(e^{nx} - e^{-nx} )$

4. anonymous

$(\sinh(x))^2 + (\cosh(x))^2 = \frac{1}{4} ( e^{x} + e^{-x} ) ^2 + \frac{1}{4}(e^{x} - e^{-x})^2$

5. anonymous

then expand , you can do the rest , its pretty easy

6. anonymous

all you need to know is (e^(x))^2 = e^(2x)

7. anonymous

then alot of things will cancel and you will get $\frac{1}{2}(e^{2x} +e^{-2x} )$

8. anonymous

which is cosh(2x) by definition