Question
If $\cos\alpha+\cos\beta=0\sin\alpha+\sin\beta,$ then prove that $\cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta).$

Answer

$\cos\alpha+\cos\beta=0\sin\alpha+\sin\beta$
Squaring on both sides gives
$\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta$
Bring square trems on one side, we get
$\cos^2\alpha+\cos^2\beta-2(-2\sin\alpha\sin\beta=\cos^2\alpha+\cos^2\beta)-2(\alpha+\beta)$

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