If ^-1x-^-1y2=, where -1 x1, -2 y2, xy2, then for all x,y, 4x^2-4xy+y^2 is equal to:

Question

If $\cos^{-1}x-\cos^{-1}\dfrac{y}{2}=\alpha$, where $-1\le x\le1$, $-2\le y\le2$, $x\le\dfrac{y}{2}$, then for all $x,y$, $4x^2-4xy\cos\alpha+y^2$ is equal to:

Accepted Answer

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

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

$$\Rightarrow4\cos^2\alpha-4xy\cos\alpha+x^2y^2=4-y^2-4x^2+x^2y^2\ \Rightarrow\ 4x^2-4xy\cos\alpha+y^2=4-4\cos^2\alpha=4\sin^2\alpha.$$

Solutions · 1