# If cos α + cos β = 0 = sin α + sin β, then prove that cos 2α + cos 2β = - 2 cos (α + β).

Proof:

cos α + cos β = 0

Squaring both sides:

(cos α + cos β)2 = (0)2

cos2 α + cos2 β + 2 cos α cos β = 0 ……(1)

sin α + sin β = 0

Squaring both sides:

(sin α + sin β)2 = (0)2

sin2 α + sin2 β + 2 sin α sin β = 0 ………(2)

Subtracting equation (1) from (2), we get

cos2 α + cos2 β + 2 cos α cos β – (sin2 α + sin2 β + 2 sin α sin β) = 0

cos2 α + cos2 β + 2 cos α cos β – sin2 α – sin2 β – 2 sin α sin β = 0

cos2 α – sin2 α + cos2 β – sin2 β + 2(cos α cos β – sin α sin β) = 0

{ cos2 x – sin2 x = 2x &

cos A cos B – sin A sin B = cos(A + B)}

cos 2α + cos 2β + 2 cos (α + β) = 0

cos 2α + cos 2β = - 2 cos (α + β)

Hence Proved

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