My Math Forum Inverse Euler's Equation

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 October 12th, 2015, 02:55 PM #1 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 Inverse Euler's Equation I am suppose to use Euler's equations to prove the following. let x represent theta in this case. $\displaystyle cos(x)cos(x)=\frac{1}{2}*cos(x+x)+\frac{1}{2}*cos( x-x)$ Would I start from the left hand side? Because I substituted cos(x) with Euler's equation and multiplied with the other cos(x). But didn't seem to be making sense. Any help would be appreciated
 October 12th, 2015, 04:23 PM #2 Newbie   Joined: Oct 2015 From: Toronto Posts: 14 Thanks: 1 The euler equation says: $cos(x) = {1\over 2}(e^{ix} +e^{-ix})$ This implies $cos(x)cos(y) ={1\over 4}(e^{ix} +e^{-ix})(e^{iy}+e^{-iy}) =$ ${1\over 4}(e^{i(x+y)} +e^{-i(x+y)}) + {1\over 4}(e^{i(x-y)}+ e^{-i(x-y)})$ $={1\over 2}(cos(x+y) + cos(x-y))$
 October 18th, 2015, 09:31 AM #3 Senior Member     Joined: Dec 2014 From: Canada Posts: 110 Thanks: 4 Thanks explains a lot

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