Question

The value of ${\sin }^3{10}^0+{\sin }^3{50}^0-{\sin }^3{70}^0$ is equal to

• A. $\frac{1}{8}$
• B. $\frac{-3}{8}$
• C. $\frac{-1}{8}$
• D. None of these

Answer 1

Answer: (B)

$\frac{-3}{8}$

Given:

${\sin }^{3}10°+{\sin }^{3}50°-{\sin }^{3}70°$

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

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

$\Rightarrow$${\sin }^{3}10°=\frac{3\sin 10°-\sin 30°}{4}$

$\Rightarrow$${\sin }^{3}50°=\frac{3\sin 50°-\sin 150°}{4}$

$\Rightarrow$${\sin }^{3}70°=\frac{3\sin 70°-\sin 210°}{4}$

$=\frac{3\sin 10°-\frac{1}{2}}{4}+\frac{3\sin (60-10)-\frac{1}{2}}{4}-\left(\frac{3\sin (60+10)+\frac{1}{2}}{4}\right)$

$=\frac{1}{4}[3\sin 10°-\frac{1}{2}+3\sin (60-10)-\frac{1}{2}-3\sin (60+10)-\frac{1}{2}]$

$=\frac{1}{4}[3\sin 10°+3\sin 60°\cos 10°-3\cos 60°\sin 10°-3\sin 60°\cos 10°-3\cos 60°\sin 10°-\frac{3}{2}]$

$=\frac{1}{4}[3\sin 10°-6\cos {60}^{\circ }.\sin {10}^{\circ }-\frac{3}{2}]$

$=\frac{1}{4}[3\sin 10°-\sin 10°-\frac{3}{2}]$

$=\frac{1}{4}\times (-\frac{3}{2})$

$=-\frac{3}{8}$