Question

If ${{\cos }^{-1}}x-{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha $ then $4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}$ is equal to?
$\begin{align}
  & \left( A \right)4{{\sin }^{2}}\alpha \\
 & \left( B \right)-4{{\sin }^{2}}\alpha \\
 & \left( C \right)2\sin 2\alpha \\
 & \left( D \right)4 \\
\end{align}$

Answer 1

Hint: We first apply the formula of ${{\cos }^{-1}}x+{{\cos }^{-1}}y$ on the left hand side of the given equation. We then simplify the equation, and square it. Rearranging the terms and applying some basic trigonometric formulae, we get $\left( A \right)$ as the correct option.

Complete step by step answer:
The given equation is
${{\cos }^{-1}}x+{{\cos }^{-1}}\left( \dfrac{y}{2} \right)=\alpha $
We know the formula that ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)$ . Thus, applying this formula in the above equation the equation thus becomes,
$\Rightarrow {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right)=\alpha $
Taking $\text{cosine}$ on both sides on the above equation, we get,
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)} \right) \right)=\cos \alpha $
We know the simple formula that $\cos \left( {{\cos }^{-1}}x \right)=\cos x$ . So, applying this in the above equation, the equation thus becomes,
$\Rightarrow x\left( \dfrac{y}{2} \right)+\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{\left( \dfrac{y}{2} \right)}^{2}} \right)}=\cos \alpha $
Simplifying the above equation, we get,
\[\Rightarrow \dfrac{xy}{2}+\sqrt{\left( 1-{{x}^{2}} \right)\left( \dfrac{4-{{y}^{2}}}{4} \right)}=\cos \alpha \]
Further simplifying the above equation, the equation thus becomes,
\[\Rightarrow \dfrac{xy}{2}+\dfrac{\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}}{2}=\cos \alpha \]
Multiplying both sides of the above equation by $2$ , we get,
\[\Rightarrow xy+\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}=2\cos \alpha \]
Taking $\cos \alpha $ to the left hand side of the above equation and \[\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}\] to the right hand side of the above equation, we get,
\[\Rightarrow xy-2\cos \alpha =-\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)}\]
Squaring both sides of the above equation, the equation thus becomes,
\[\Rightarrow {{\left( xy-2\cos \alpha \right)}^{2}}={{\left( -\sqrt{\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)} \right)}^{2}}\]
Evaluating the above equation, we get
\[\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =\left( 1-{{x}^{2}} \right)\left( 4-{{y}^{2}} \right)\]
Opening the brackets in the above equation, we get,
\[\Rightarrow {{x}^{2}}{{y}^{2}}-4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}\]
Subtracting \[{{x}^{2}}{{y}^{2}}\] from both sides of the above equation, we get,
\[\Rightarrow -4xy\cos \alpha +4{{\cos }^{2}}\alpha =4-4{{x}^{2}}-{{y}^{2}}\]
Bringing the terms $4{{x}^{2}},{{y}^{2}}$ to the left hand side of the above equation and the term \[4{{\cos }^{2}}\alpha \] to the right hand side of the above equation, we get,
\[\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4-4{{\cos }^{2}}\alpha \]
Taking $4$ common in the right hand side of the above equation, we get,
\[\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4\left( 1-{{\cos }^{2}}\alpha \right)\]
We know that \[1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha \] . Thus, applying this formula in the above equation, we get,
\[\Rightarrow 4{{x}^{2}}+{{y}^{2}}-4xy\cos \alpha =4{{\sin }^{2}}\alpha \]
Rearranging the terms of the above equation, we get,
\[\Rightarrow 4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha \]
This is nothing but the thing that we have to prove. Therefore, we can conclude that \[4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}\] is equal to \[4{{\sin }^{2}}\alpha \] which is option $\left( A \right)$ .

Note: We must be very careful while carrying out the square as this expression deals with a little complex terms and students are prone to make mistakes here. This problem can also be solved by taking some values for $x,y,\alpha $ and find out which of the following options gives the correct answer. Let’s take $x=0,y=1$ . Then, $\alpha $ becomes ${{\cos }^{-1}}0-{{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}-\dfrac{\pi }{3}=\dfrac{\pi }{6}$ . The expression becomes $4{{\left( 0 \right)}^{2}}-4\left( 0 \right)\left( 1 \right)\cos \left( \dfrac{\pi }{6} \right)+{{\left( 1 \right)}^{2}}=1$ . Out of the following options, only $\left( A \right)$ satisfies by putting $\alpha =\dfrac{\pi }{6}$.