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
45