Question

If $\sin A+\sin 2A=x$ and $\cos A+\cos 2A=y$, then $\left(x^2+y^2\right)\left(x^2+\right.\left.y^2-3\right)=$

• A. $2y$
• B. $y$
• C. $3y$
• D. None of these

Answer 1

The correct option is A

$2y$

Explanation for the correct option.

Step 1: Find the value of $\left(x^2+y^2\right)$

Given that, $\sin A+\sin 2A=x$ and $\cos A+\cos 2A=y$.

$\begin{array}{rcl}\therefore \left(x^2+y^2\right)& =& {\left(\sin A+\sin 2A\right)}^2+{\left(\cos A+\cos 2A\right)}^2\\ & =& {\sin }^{2}A+{\sin }^{2}2A+2\sin 2A\sin A+{\cos }^{2}A+{\cos }^{2}2A+2\cos A\cos 2A\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right]\\ & =& \left({\sin }^{2}A+{\cos }^{2}A\right)+\left(si{n}^{2}2A+{\cos }^{2}2A\right)+2\left(\sin A\sin 2A+\cos A\cos 2A\right)\\ & =& 1+1+2\cos \left(2A-A\right)\\ & =& 2(1+\cos A).....\left(1\right)\end{array}$

Step 2: Find the value of $\left(x^2+\right.\left.y^2-3\right)$

By (1) we have:

$\begin{array}{rcl}\therefore \left(x^2+\right.\left.y^2\right)\left(x^2+\right.\left.y^2-3\right)& =& 2\left(1+\cos A\right)\left[2\left(1+\cos A\right)-3\right]\\ & =& 2\left(1+\cos A\right)\left(2+2\cos A-3\right)\\ & =& \left(2+2\cos A\right)\left(2\cos A-1\right)\\ & =& 4\cos A-2+4{\cos }^{2}A-2\cos A\\ & =& 2\cos A+4{\cos }^{2}A-2\\ & =& 2\left(2{\cos }^2\mathrm{A}-1\right)+2\mathrm{cosA}\\ & =& 2\cos 2\mathrm{A}+2\mathrm{cosA}\left[\because \cos 2A=2{\cos }^{2}A-1\right]\\ & =& 2(\mathrm{cosA}+\cos 2\mathrm{A})\\ & =& 2\mathrm{y}\left\{\text{from the given}\right\}\end{array}$

Therefore, $\left(x^2+y^2\right)\left(x^2+\right.\left.y^2-3\right)=2y$

Hence, option A is correct.