I tried starting with the left part of the equation which did not lead anywhere near the second
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Use $e^{ix}=$cos$(x)+i$sin$(x)$ – Nightgap Nov 26 '17 at 01:02
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Use https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b – lab bhattacharjee Nov 26 '17 at 11:18
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Hint
Here are the steps $$\begin{align}\sin^2(a+b)&=(\sin a\cos b+\sin b\cos a)^2\\&=\sin^2a\cos^2b+\sin^2b\cos^2a+2\sin a\cos b\sin b\cos a\\&=\sin^2a(1-\sin^2b)+\sin^2b(1-\sin^2a)+2\sin a\cos b\sin b\cos a\\&=\sin^2a+\sin^2b+2(\sin a\cos b\sin b\cos a-\sin^2a\sin^2b)\end{align}$$ Can you proceed from here?
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Assuming you know the formulas for $\sin(a+b)$ and $\cos(a+b)$:
\begin{align} \sin^2(a+b) &= \sin^2 a \cos^2 b + \cos^2 a \sin^2 b + 2\sin a \sin b \cos a \cos b \\ &= \sin^2 a (1-\sin^2b)+(1-\sin^2a)\sin^2b + 2\sin a \sin b \cos a \cos b\\ &= \sin^2a +\sin^2b + 2\sin a\sin b(\cos a\cos b-\sin a\sin b) \\ &=\sin^2a +\sin^2b+2\sin a\sin b \cos(a+b) \end{align}