For some reason I cannot get the solution.
Show that $x$ is independent of: $\sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z)$
I have used all the identities but seem to be missing something.
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1Hint: do you know the law of cosines? – symplectomorphic Jan 16 '15 at 18:38
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Yea the one you use in 2D and 3D problems or another one? – Jared Riley Jan 16 '15 at 18:40
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$\begin{align} & \sin^2(x+y)+\sin^2(x+z)-2\cos(y-z)\sin(x+y)\sin(x+z) \\ &=\dfrac{1}{2}(1-\cos(2x+2y)) + \dfrac{1}{2}(1-\cos(2x+2z)) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\ &=1-\cos(2x+y+z)\cos(y-z) - 2\cos(y-z)\sin(x+y)\sin(x+z)\\ &=1-\cos(y-z)\left[\cos(2x+y+z) + 2\sin(x+y)\sin(x+z) \right]\\ &=1-\cos(y-z)\left[\cos(2x+y+z) + \cos(y-z) - \cos(2x+y+z)\right]\\ &=1-\cos^2(y-z)\\ \end{align}$
which is independent of $x.$
abel
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$\sin^2(x+y)+\sin^2(x+z) =1-[\cos^2(x+y)-\sin^2(x+z)]$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$\cos^2(x+y)-\sin^2(x+z)=\cos(2x+y+z)\cos(y-z)$
and then use Werner formula, for $2\sin(x+y)\sin(x+z)$
lab bhattacharjee
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