If cos (α + β) = 0, then sin (α – β) can be reduced to (A) cos β (B) cos 2β (C) sin α (D) sin 2α
If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α
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(B) cos 2β
According to the question,
cos(α+β) = 0
Since, cos 90° = 0
We can write,
cos(α+β)= cos 90°
By comparing cosine equation on L.H.S and R.H.S,
We get,
(α+β)= 90°
α = 90°-β
Now we need to reduce sin (α -β ),
So, we take,
sin(α-β) = sin(90°-β-β) = sin(90°-2β)
sin(90°-θ) = cos θ
So, sin(90°-2β) = cos 2β
Therefore, sin(α-β) = cos 2β
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