Question:
If cosα + cosβ = 0 = sinα + sinβ, then prove that cos 2α + cos 2β = –2cos (α + β).
[Hint: cosα + cosβ)2 – (sinα + sinβ)2 = 0]
Solution:
According to the question,
cosα + cosβ = 0 = sinα + sinβ …(i)
Since, LHS = cos 2α + cos 2β
We know that,
cos 2x = cos2x – sin2x
Therefore,
LHS = cos2α – sin2α + (cos2β – sin2β)
⇒ LHS = cos2α + cos2β – (sin2α + sin2β)
Also, since,
a2 + b2 = (a+b)2 – 2ab
⇒ LHS = (cosα + cosβ)2 – 2cosα cosβ –(sinα + sinβ)2 +2sinα sinβ
From equation (i),
⇒ LHS = 0 – 2cosα cosβ -0 + 2sinα sinβ
⇒ LHS = -2(cosα cosβ – sinα sinβ)
∵ cos (α + β) = cosα cosβ – sinα sinβ
Therefore, LHS = -2 cos (α + β) = RHS
Hence, cos 2α + cos 2β = –2cos (α + β)