Question 1116833
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One term of the LHS is *[tex \Large \sin^2(a\,+\,b)] and the other is *[tex \Large \cos^2(a\,+\,b)].  So let *[tex \Large (a\,+\,b)\ =\ \varphi], and *[tex \Large \sin^2\varphi\ +\ \cos^2\varphi\ \equiv\ 1], Q.E.D


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  


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