If cos⁡α+cos⁡β=0=sin⁡α+sin⁡β then cos⁡2α+cos⁡2β is equal to

A
$2 \cos (α + β)$
B
$- 2 \cos (α + β)$
C
$3 \cos (α + β)$
D
None of these

Solution:

Given $\cos α + \cos β = 0 and \sin α + \sin β = 0$

On squaring and subtracting both the equations, we get

$\begin{array}{ll} (\cos α + \cos β)^{2} - (\sin α + \sin β)^{2} = 0 \\ \Rightarrow & (\cos^{2} α + \cos^{2} β + 2 \cos αcos β) - (\sin^{2} α + \sin^{2} β + 2 \sin αsin β) = 0 \\ \Rightarrow & (\cos^{2} α - \sin^{2} α) + (\cos^{2} β - \sin^{2} β) + 2 [\cos αcos β - \sin αsin β] = 0 \\ \Rightarrow & \cos 2 α + \cos 2 β + 2 \cos (α + β) = 0 \\ \Rightarrow & \cos 2 α + \cos 2 β = - 2 \cos (α + β) \end{array}$

If $\cos α + \cos β = 0 = \sin α + \sin β$ then $\cos 2 α + \cos 2 β$ is equal to

Solution:

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