Question

If $x+y=3-\cos 4\theta$ and $x-y=4\sin 2\theta$ then prove that $x^2+y^2-8x-8y-2xy+16=0$

Answer 1

Proof:

It is given that,

$$\begin{gathered} x+y=3-\cos 4\theta \\ \Rightarrow x+y=3-\left(1-2{\sin }^{2}2\theta \right)\left[\because \cos 2\theta =1-2{\sin }^2\theta \right] \\ \Rightarrow x+y=2+2{\sin }^{2}2\theta ...\left(1\right) \end{gathered}$$

$$\begin{gathered} \because x-y=4\sin 2\theta \\ \therefore \sin 2\theta =\frac{x-y}{4}...\left(2\right) \end{gathered}$$

Substitute the value of the equation $\left(2\right)$ into the equation $\left(1\right)$.

$$\begin{gathered} \Rightarrow x+y=2+2{\left(\frac{x-y}{4}\right)}^2 \\ \Rightarrow x+y=2+2\left(\frac{x^2+y^2-2xy}{16}\right) \\ \Rightarrow x+y=2+\left(\frac{x^2+y^2-2xy}{8}\right) \\ \Rightarrow x+y=\frac{x^2+y^2-2xy+16}{8} \\ \Rightarrow 8x+8y=x^2+y^2-2xy+16(\mathrm{Cross}\mathrm{multiply}) \\ \therefore x^2+y^2-8x-8y-2xy+16=0 \end{gathered}$$

Hence, proved.