Question

$\frac{{\sin }^{3}3A}{{\sin }^{2}A}-\frac{{\cos }^{2}3A}{{\cos }^{2}A}=$

• A. $\frac{1}{8\cos 2A}$
• B. 8 cos 2A
• C. cos 2A
• D. none of these

Answer 1

Answer: (B)

$\frac{{\sin }^{2}3A}{{\sin }^{2}A}-\frac{{\cos }^{2}3A}{{\cos }^{2}A}$
$={\left(\frac{3\sin A-4{\sin }^{3}A}{\sin A}\right)}^2-{\left(\frac{4{\cos }^{3}A-3\cos A}{\cos A}\right)}^2$
$={\left(3-4{\sin }^{2}A\right)}^2-{\left(4{\cos }^{2}A-3\right)}^2$
$=16{\sin }^{4}A-16{\cos }^{4}A-24\left({\sin }^{2}A-{\cos }^{2}A\right)$
$=16\left({\sin }^{2}A-{\cos }^{2}A\right)\left({\sin }^{2}A+{\cos }^{2}A\right)-24\left({\sin }^{2}A-{\cos }^{2}A\right)$
$=-8\left({\sin }^{2}A-{\cos }^{2}A\right)=8\cos 2A$