If cos (θ-α)=a, sin (θ-β)=b, then cos 2(α-β)+2 a b sin (α-β) is equal to

Question

If cos (θ-α)=a, sin (θ-β)=b, then cos 2(α-β)+2 a b sin (α-β) is equal to (A) 4 a2 b2 (B) a2-b2 (C) a2+b2 (D) -a2 b2

Tardigrade Admin · Accepted Answer

We have sin (α-β)= sin (θ-β- overlineθ-α) = sin (θ-β) cos (θ-α)- cos (θ-β) sin (θ-α) =b a-√1-b2 √1-a2 text and cos (α-β)= cos (θ-β- overlineθ-α) = cos (θ-β) cos (θ-α)+ sin (θ-β) sin (θ-α) =a √1-b2+b √1-a2 ∴ text Given expression is cos 2(α-β)+2 a b sin (α-β) a prime =a2+b2

Q. If $cos (θ - α) = a, sin (θ - β) = b$ , then $cos^{2} (α - β) + 2 ab sin (α - β)$ is equal to

$4 a^{2} b^{2}$

$a^{2} - b^{2}$

$a^{2} + b^{2}$

$- a^{2} b^{2}$

Q. If cos(θ−α)=a,sin(θ−β)=b, then cos2(α−β)+2absin(α−β) is equal to

4a2b2

a2−b2

a2+b2

−a2b2

Q. If $cos (θ - α) = a, sin (θ - β) = b$ , then $cos^{2} (α - β) + 2 ab sin (α - β)$ is equal to

$4 a^{2} b^{2}$

$a^{2} - b^{2}$

$a^{2} + b^{2}$

$- a^{2} b^{2}$