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Prove cosh(2x) = cosh^(2)x + sin^(2)x
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its just going back to the definitions
\[\cosh (nx) = \frac{1}{2} ( e^{nx} + e^{-nx} ) \]
\[\sinh(nx) = \frac{1}{2}(e^{nx} - e^{-nx} )\]
\[(\sinh(x))^2 + (\cosh(x))^2 = \frac{1}{4} ( e^{x} + e^{-x} ) ^2 + \frac{1}{4}(e^{x} - e^{-x})^2\]
then expand , you can do the rest , its pretty easy
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all you need to know is (e^(x))^2 = e^(2x)
then alot of things will cancel and you will get \[\frac{1}{2}(e^{2x} +e^{-2x} ) \]
which is cosh(2x) by definition
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