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Mathematics Question on Trigonometric Functions

If sin⁡α+sin⁡β=0=cos⁡α+cos⁡β\sin \alpha + \sin \beta = 0 = \cos \alpha + \cos \beta, then cos⁡2α+cos⁡2β=\cos 2\alpha + \cos 2\beta =

A

−2sin⁡(α+β)- 2 \sin (\alpha + \beta)

B

−2cos⁡(α+β)- 2 \cos (\alpha + \beta)

C

2sin⁡(α+β)2 \sin (\alpha + \beta)

D

2cos⁡(α+β)2 \cos (\alpha + \beta)

Answer

−2cos⁡(α+β)- 2 \cos (\alpha + \beta)

Explanation

Solution

Given cos⁡α+cos⁡β=0 \cos\alpha +\cos\beta = 0 ....(1) and sin⁡α+sin⁡β=0\sin\alpha + \sin\beta = 0 ....(2) Squaring (1) and (2) and subtracting, we get (cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2=0\left(\cos\alpha + \cos \beta\right)^{2} - \left(\sin\alpha +\sin\beta\right)^{2} = 0 (cos⁡2α−sin⁡2α)+(cos⁡2β−sin⁡2β)+2(cos⁡αcos⁡β−sin⁡αsin⁡β)=0\left(\cos^{2} \alpha -\sin^{2}\alpha\right) +\left(\cos^{2}\beta -\sin^{2} \beta \right) + 2 \left(\cos\alpha \cos\beta -\sin\alpha \sin\beta \right) = 0 or cos⁡2α+cos⁡2β=−2cos⁡(α+β)\cos2 \alpha +\cos2\beta = - 2 \cos\left(\alpha +\beta \right)