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Question: If \(\sin \left( \theta +\alpha \right)=a\) and \(\sin \left( \theta +\beta \right)=b\) , \(\left( \...

If sin⁡(θ+α)=a\sin \left( \theta +\alpha \right)=a and sin⁡(θ+β)=b\sin \left( \theta +\beta \right)=b , ((0<α,β,θ<π2))\left( \left( 0<\alpha ,\beta ,\theta <\dfrac{\pi }{2} \right) \right) then 2cos⁡2(α−β)−1−4abcos⁡(α−β)2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right) is
(A)1−a2−b21-{{a}^{2}}-{{b}^{2}}
(B)1−2a2−2b21-2{{a}^{2}}-2{{b}^{2}}
(C)2+a2+b22+{{a}^{2}}+{{b}^{2}}
(D)2−a2−b22-{{a}^{2}}-{{b}^{2}}

Explanation

Solution

For answering this question we will first find the value of cos⁡(θ+α)\cos \left( \theta +\alpha \right) and cos⁡(θ+β)\cos \left( \theta +\beta \right) using the values of sin⁡(θ+α)=a\sin \left( \theta +\alpha \right)=a and sin⁡(θ+β)=b\sin \left( \theta +\beta \right)=b from the formulae cos⁡x=1−sin⁡2x\cos x=\sqrt{1-{{\sin }^{2}}x}. And after that we will find the value of cos⁡(α−β)=cos⁡[(θ+α)−(θ+β)]\cos \left( \alpha -\beta \right)=\cos \left[ \left( \theta +\alpha \right)-\left( \theta +\beta \right) \right] using cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B and substitute it in the equation we have 2cos⁡2(α−β)−1−4abcos⁡(α−β)2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right) and simplify it.

Complete step by step answer:
From the question we have that the value of sin⁡(θ+α)=a\sin \left( \theta +\alpha \right)=a and sin⁡(θ+β)=b\sin \left( \theta +\beta \right)=b.
Since we know that cos⁡x=1−sin⁡2x\cos x=\sqrt{1-{{\sin }^{2}}x} we can say that the value of cos⁡(θ+α)=1−a2\cos \left( \theta +\alpha \right)=\sqrt{1-{{a}^{2}}} andcos⁡(θ+β)=1−b2\cos \left( \theta +\beta \right)=\sqrt{1-{{b}^{2}}} .
Here we need the value of cos⁡(α−β)\cos \left( \alpha -\beta \right) in order to answer the question. We can write α−β\alpha -\beta as (θ+α)−(θ+β)\left( \theta +\alpha \right)-\left( \theta +\beta \right) by adding and subtracting θ\theta . Since we know that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B . We can have the value of cos⁡(α−β)\cos \left( \alpha -\beta \right) as cos⁡(α−β)=cos⁡[(θ+α)−(θ+β)]=cos⁡(θ+α)cos⁡(θ+β)+sin⁡(θ+α)sin⁡(θ+β)\cos \left( \alpha -\beta \right)=\cos \left[ \left( \theta +\alpha \right)-\left( \theta +\beta \right) \right]=\cos \left( \theta +\alpha \right)\cos \left( \theta +\beta \right)+\sin \left( \theta +\alpha \right)\sin \left( \theta +\beta \right) .
By substituting the respective values we will have 1−a21−b2+ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab .
By substituting this value cos⁡(α−β)=1−a21−b2+ab\cos \left( \alpha -\beta \right)=\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab in the place of the required value we will get 2cos⁡2(α−β)−1−4abcos⁡(α−β)=2(1−a21−b2+ab)2−1−4ab(1−a21−b2+ab)2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=2{{\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)}^{2}}-1-4ab\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right) .
By simplifying this equation we will have 2(1−a21−b2+ab)2−1−4a2b2−4ab1−a21−b22{{\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)}^{2}}-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} .
Since we know that (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab by using this we will have 2((1−a2)(1−b2)+2ab(1−a21−b2)+a2b2)−1−4a2b2−4ab1−a21−b22\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right)+2ab\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} \right)+{{a}^{2}}{{b}^{2}} \right)-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} .
By simplifying this we will have 2((1−a2)(1−b2))+2a2b2+4ab1−a21−b2−1−4a2b2−4ab1−a21−b2 ⇒2((1−a2)(1−b2))−1−2a2b2 \begin{aligned} & 2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)+2{{a}^{2}}{{b}^{2}}+4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} \\\ & \Rightarrow 2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)-1-2{{a}^{2}}{{b}^{2}} \\\ \end{aligned} .
By expanding this we will have 2((1−a2)(1−b2))−1−2a2b2⇒2(1−a2−b2+a2b2)−1−2a2b22\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)-1-2{{a}^{2}}{{b}^{2}}\Rightarrow 2\left( 1-{{a}^{2}}-{{b}^{2}}+{{a}^{2}}{{b}^{2}} \right)-1-2{{a}^{2}}{{b}^{2}}.
By expanding and simplifying this we will have 2−2a2−2b2+2a2b2−1−2a2b2=1−2a2−2b22-2{{a}^{2}}-2{{b}^{2}}+2{{a}^{2}}{{b}^{2}}-1-2{{a}^{2}}{{b}^{2}}=1-2{{a}^{2}}-2{{b}^{2}}.
Hence we end up with the value that 2cos⁡2(α−β)−1−4abcos⁡(α−β)=1−2a2−2b22{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=1-2{{a}^{2}}-2{{b}^{2}} .
So now we have a conclusion that when sin⁡(θ+α)=a\sin \left( \theta +\alpha \right)=a and sin⁡(θ+β)=b\sin \left( \theta +\beta \right)=b , then 2cos⁡2(α−β)−1−4abcos⁡(α−β)=1−2a2−2b22{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=1-2{{a}^{2}}-2{{b}^{2}}.

So, the correct answer is “Option B”.

Note: While answering of questions of this type we should remember that the formula is cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B not cos⁡(A−B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B-\sin A\sin B . If we use this we will end up having a wrong answer.