Question

$32{\sin }^6{15}^{\circ }-48{\sin }^4{15}^{\circ }+18{\sin }^2{15}^{\circ }=$

• A. $1$
• B. $2$
• C. $3$
• D. $-1$

Answer 1

Answer: (A)

Recall that,

$\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

Put $\theta ={15}^{\circ }$

$\Rightarrow \sin (3\times {15}^{\circ })=3\sin {15}^{\circ }-4{\sin }^3{15}^{\circ }$

$\Rightarrow \sin {45}^{\circ }=3\sin {15}^{\circ }-4{\sin }^3{15}^{\circ }$

$\Rightarrow \frac{1}{\sqrt{2}}=3\sin {15}^{\circ }-4{\sin }^3{15}^{\circ }$

Now squaring both sides we have,

$\Rightarrow (\frac{1}{\sqrt{2}}{)}^2=(3\sin {15}^{\circ }-4{\sin }^3{15}^{\circ }{)}^2$

$\Rightarrow \frac{1}{2}=3^2{\sin }^2{15}^{\circ }-2\left(3\sin {15}^{\circ }\right)\left(4{\sin }^3{15}^{\circ }\right)+4^2({\sin }^3{15}^{\circ }{)}^2$

$\Rightarrow \frac{1}{2}=9{\sin }^2{15}^{\circ }-24{\sin }^4{15}^{\circ }+16{\sin }^6{15}^{\circ }$

$\Rightarrow 1=18{\sin }^2{15}^{\circ }-48{\sin }^4{15}^{\circ }+32{\sin }^6{15}^{\circ }$

$\therefore 32{\sin }^6{15}^{\circ }-48{\sin }^4{15}^{\circ }+18{\sin }^2{15}^{\circ }=1$

Hence, Option A is correct.