Question

${\cos }^2\left(A\right){[3-4{\cos }^2\left(A\right)]}^2+{\sin }^2\left(A\right){[3-4{\sin }^2\left(A\right)]}^2=$

• A. $\cos \left(4A\right)$
• B. $\sin \left(4A\right)$
• C. $1$
• D. None of these.

Answer 1

The correct option is C

$1$

Simplify the series using Trigonometric identity:

$\begin{array}{rcl}& & {\cos }^2\left(A\right){[3-4{\cos }^2\left(A\right)]}^2+{\sin }^2\left(A\right){[3-4{\sin }^2\left(A\right)]}^2\\ & =& {\left\{\cos A\right(3-4{\cos }^{2}A\left)\right\}}^2+{\left\{\sin A\right(3-4{\sin }^{2}A\left)\right\}}^2\\ & =& {\{3\cos A-4{\cos }^{3}A\}}^2+{\{3\sin A-4{\sin }^{3}A\}}^2\\ & =& \{\cos 3A{\}}^2+{\left\{\sin 3A\right\}}^2\{\because \cos 3A=4{\cos }^{3}A-3\cos A,\sin 3A=3\sin A-4{\sin }^{3}A\}\\ & =& {\cos }^{2}3A+{\sin }^{2}3A\{\because {\cos }^{2}A+{\sin }^{2}A=1)\\ & =& 1\end{array}$

Therefore, Option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is the correct answer.