Question

If ${\cos }^{-1}x-{\cos }^{-1}\left(\frac{y}{2}\right)=\alpha$, then $4x^2-4xy\cos \alpha +y^2=$

• A. $4{\sin }^2\alpha$
• B. $-4{\sin }^2\alpha$
• C. $2\sin 2\alpha$
• D. $4$

Answer 1

The correct option is A

$4{\sin }^2\alpha$

Explanation for the correct option:

Step 1: Apply inverse trigonometric identity.

We have given,

${\cos }^{-1}x-{\cos }^{-1}\left(\frac{y}{2}\right)=\alpha$,

we know that,

$\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{-}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{\{}\mathit{x}\mathit{y}\mathbf{+}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{.}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\mathbf{\}}$

by using the above identity,

${\cos }^{-1}\{\frac{xy}{2}+\sqrt{1-x^2}.\sqrt{1-\frac{y^2}{4}}\}=\alpha$

$\Rightarrow$ $\frac{xy}{2}+\sqrt{1-x^2}.\sqrt{1-\frac{y^2}{4}}=\cos \alpha$

$\Rightarrow$ $\sqrt{1-x^2}.\sqrt{1-\frac{y^2}{4}}=\cos \alpha -\frac{xy}{2}$

Step 2: Simplifying and squaring both sides,

$\begin{array}{rcl}{(2\sqrt{1-x^2}.\sqrt{1-\frac{y^2}{4}})}^2& =& {(2\cos \alpha -xy)}^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{4(1-x^2)(4-y^2)}{4}& =& 4{\cos }^2\alpha +x^2{y}^2-4xy\cos \alpha \end{array}$

$\Rightarrow$$\begin{array}{rcl}4-y^2-4x^2+x^2{y}^2+4xy\cos \alpha -x^2{y}^2& =& 4{\cos }^2\alpha \end{array}$

$\Rightarrow$ $\begin{array}{rcl}4-4x^2-y^2+4xy\cos \alpha & =& 4{\cos }^2\alpha \end{array}$

$\Rightarrow$ $\begin{array}{rcl}4-4{\cos }^2\alpha & =& 4x^2+y^2-4xy\cos \alpha \end{array}$

$\Rightarrow$ $\begin{array}{rcl}4{\sin }^2\alpha & =& 4x^2+y^2-4xy\cos \alpha \end{array}$

Hence, the correct option is$\mathbf{\left(}\mathit{A}\mathbf{\right)}$.