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Question
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
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Solution
Given: \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]
\[\cos^2 \alpha + \cos^2 \beta + 2\cos\alpha \cos\beta = 0 . . . . . \left( 1 \right)\]
Also,
\[\sin\alpha + \sin\beta = 0\]
\[ \Rightarrow \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + 2\left( \cos\alpha \cos\beta - \sin\alpha \sin\beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta + 2\cos\left( \alpha + \beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]
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