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Question: If we have an expression as \({{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha \), where \(-1\le x\...

If we have an expression as cos⁡−1x−cos⁡−1y2=α{{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha , where −1≤x≤1,−2≤y≤2,x≤y2-1\le x\le 1,-2\le y\le 2,x\le \dfrac{y}{2} the 4x2−4xycos⁡α+y24{{x}^{2}}-4xy\cos \alpha +{{y}^{2}} is equal to
[a] 4sin⁡2α+2x2y24{{\sin }^{2}}\alpha +2{{x}^{2}}{{y}^{2}}
[b] 4sin⁡2α−2x2y24{{\sin }^{2}}\alpha -2{{x}^{2}}{{y}^{2}}
[c] 4sin⁡2α4{{\sin }^{2}}\alpha
[d] None of these

Explanation

Solution

Take cos on both sides of the equation cos⁡−1x−cos⁡−1y2=α{{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha and use the fact that cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B. Simplify and hence find the value of the expression 4x2−4xycos⁡α+y24{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}.

Complete step-by-step solution:
We have cos⁡−1x−cos⁡−1y2=α{{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}=\alpha
Taking cos on both sides, we get
cos⁡(cos⁡−1x−cos⁡−1y2)=cos⁡α\cos \left( {{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2} \right)=\cos \alpha
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
Hence, we have
cos⁡(cos⁡−1x)cos⁡(cos⁡−1y2)+sin⁡(cos⁡−1x)sin⁡(cos⁡−1y2)=cos⁡α\cos \left( {{\cos }^{-1}}x \right)\cos \left( {{\cos }^{-1}}\dfrac{y}{2} \right)+\sin \left( {{\cos }^{-1}}x \right)\sin \left( {{\cos }^{-1}}\dfrac{y}{2} \right)=\cos \alpha
We know that cos⁡(cos⁡−1x)=x∀x∈[−1,1]\cos \left( {{\cos }^{-1}}x \right)=x\forall x\in \left[ -1,1 \right] and sin⁡(cos⁡−1x)=1−x2∀x∈[−1,1]\sin \left( {{\cos }^{-1}}x \right)=\sqrt{1-{{x}^{2}}}\forall x\in \left[ -1,1 \right]
Hence, we have
xy2+1−x2×1−y24=cos⁡αx\dfrac{y}{2}+\sqrt{1-{{x}^{2}}}\times \sqrt{1-\dfrac{{{y}^{2}}}{4}}=\cos \alpha
Subtracting xy2\dfrac{xy}{2} on both sides, we get
1−x24−y22=cos⁡α−xy2\sqrt{1-{{x}^{2}}}\dfrac{\sqrt{4-{{y}^{2}}}}{2}=\cos \alpha -\dfrac{xy}{2}
Multiplying by 2 on both sides, we get
1−x24−y2=2cos⁡α−xy\sqrt{1-{{x}^{2}}}\sqrt{4-{{y}^{2}}}=2\cos \alpha -xy
Squaring both sides, we get
(1−x2)(4−y2)=(2cos⁡α−xy)2\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)={{\left( 2\cos \alpha -xy \right)}^{2}}
We know that (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}
Hence, we have
(1−x2)(4−y2)=4cos⁡2α+x2y2−4xycos⁡α\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)=4{{\cos }^{2}}\alpha +{{x}^{2}}{{y}^{2}}-4xy\cos \alpha
Expanding the term on LHS, we get
4−y2−4x2+x2y2=4cos⁡2α−4xycos⁡α+x2y24-{{y}^{2}}-4{{x}^{2}}+{{x}^{2}}{{y}^{2}}=4{{\cos }^{2}}\alpha -4xy\cos \alpha +{{x}^{2}}{{y}^{2}}
Subtracting x2y2{{x}^{2}}{{y}^{2}} from both sides, we get
4−y2−4x2=4cos⁡2α−4xycos⁡α4-{{y}^{2}}-4{{x}^{2}}=4{{\cos }^{2}}\alpha -4xy\cos \alpha
Adding 4x2+y24{{x}^{2}}+{{y}^{2}} on both sides, we get
4=4x2+y2−4xycos⁡α+4cos⁡2α4=4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha
Subtracting 4cos⁡2α4{{\cos }^{2}}\alpha on both sides, we get
4x2−4xycos⁡α+y2=4−4cos⁡2α=4(1−cos⁡2α)4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4-4{{\cos }^{2}}\alpha =4\left( 1-{{\cos }^{2}}\alpha \right)
We know that 1−cos⁡2α=sin⁡2α1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha
Hence, we have
4x2−4xycos⁡α+y2=4sin⁡2α4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha
Hence option [c] is correct.

Note: Alternative solution:
We know that cos⁡−1x−cos⁡−1y=cos⁡−1(xy−1−x21−y2)+2kπ,k∈N{{\cos }^{-1}}x-{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)+2k\pi ,k\in \mathbb{N}, where k is suitably chosen.
Hence, we have
cos⁡−1x−cos⁡−1y2=cos⁡−1(xy2−1−x21−y24)+2kπ,k∈N ⇒α=cos⁡−1(xy2−1−x21−y24)+2kπ,k∈N \begin{aligned} & {{\cos }^{-1}}x-{{\cos }^{-1}}\dfrac{y}{2}={{\cos }^{-1}}\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right)+2k\pi ,k\in \mathbb{N} \\\ & \Rightarrow \alpha ={{\cos }^{-1}}\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right)+2k\pi ,k\in \mathbb{N} \\\ \end{aligned}
Taking cos on both sides, we get
cos⁡α=(xy2−1−x21−y24)\cos \alpha =\left( \dfrac{xy}{2}-\sqrt{1-{{x}^{2}}}\sqrt{1-\dfrac{{{y}^{2}}}{4}} \right), which is the same as obtained above. Hence proceeding as above, we get option [c] is correct.