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Question: If \(\sin(\theta + \alpha) = a\) and \(\sin(\theta + \beta) = b,\) then \(\cos 2(\alpha - \beta) - ...

If sin⁡(θ+α)=a\sin(\theta + \alpha) = a and sin⁡(θ+β)=b,\sin(\theta + \beta) = b, then

cos⁡2(α−β)−4abcos⁡(α−β)\cos 2(\alpha - \beta) - 4ab\cos(\alpha - \beta) is equal to

A

1−a2−b21 - a^{2} - b^{2}

B

1−2a2−2b21 - 2a^{2} - 2b^{2}

C

2+a2+b22 + a^{2} + b^{2}

D

2−a2−b22 - a^{2} - b^{2}

Answer

1−2a2−2b21 - 2a^{2} - 2b^{2}

Explanation

Solution

Given that sin⁡(θ+α)=a\sin(\theta + \alpha) = a …..(i)

and sin⁡(θ+β)=b\sin(\theta + \beta) = b …..(ii)

Now, cos⁡(θ+α)=1−a2⇒θ+α=cos⁡−11−a2\cos(\theta + \alpha) = \sqrt{1 - a^{2}} \Rightarrow \theta + \alpha = \cos^{- 1}\sqrt{1 - a^{2}}

and α−β=(θ+α)−(θ+β)\alpha - \beta = (\theta + \alpha) - (\theta + \beta)

=cos⁡−11−a2−cos⁡−11−b2= \cos^{- 1}\sqrt{1 - a^{2}} - \cos^{- 1}\sqrt{1 - b^{2}}

⇒α−β=cos⁡−1(1−a21−b2+ab)\Rightarrow \alpha - \beta = \cos^{- 1}(\sqrt{1 - a^{2}}\sqrt{1 - b^{2}} + ab)

⇒cos⁡(α−β)=1−a21−b2+ab\Rightarrow \cos(\alpha - \beta) = \sqrt{1 - a^{2}}\sqrt{1 - b^{2}} + ab

Now, cos⁡2(α−β)−4abcos⁡(α−β)\cos{}2(\alpha - \beta) - 4ab\cos(\alpha - \beta)

=2cos⁡2(α−β)−1−4abcos⁡(α−β)= 2\cos^{2}{}(\alpha - \beta) - 1 - 4ab\cos(\alpha - \beta)

=2(1−a21−b2+ab)2−4ab(1−a21−b2+ab)−1= 2\left( \sqrt{1 - a^{2}}\sqrt{1 - b^{2}} + ab \right)^{2} - 4ab\left( \sqrt{1 - a^{2}}\sqrt{1 - b^{2}} + ab \right) - 1

=2{(1−a2)(1−b2)+a2b2+2ab1−a21−b2}= 2\{(1 - a^{2})(1 - b^{2}) + a^{2}b^{2} + 2ab\sqrt{1 - a^{2}}\sqrt{1 - b^{2}}\}

−4ab(1−a21−b2+ab)- 4ab(\sqrt{1 - a^{2}}\sqrt{1 - b^{2}} + ab)

=2(1−b2−a2+a2b2)+2a2b2−4a2b2−1= 2(1 - b^{2} - a^{2} + a^{2}b^{2}) + 2a^{2}b^{2} - 4a^{2}b^{2} - 1

=2(1−a2−b2)−1=1−2a2−2b2.= 2(1 - a^{2} - b^{2}) - 1 = 1 - 2a^{2} - 2b^{2}.