Question

The value of ${\sin }^3{10}^o+{\sin }^3{50}^o-{\sin }^3{70}^o=$

• A. $-\frac{3}{2}$
• B. $\frac{3}{4}$
• C. $-\frac{3}{4}$
• D. $-\frac{3}{8}$

Answer 1

Answer: (D)

$-\frac{3}{8}$

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

$\therefore E=\frac{1}{4}[(3\sin {10}^o-\sin {30}^o)+(3\sin {50}^o-\sin {150}^o)-(3\sin {70}^o-\sin {210}^o)]$

$=\frac{1}{4}[3(\sin {10}^o+\sin {50}^o-\sin {70}^o)-\{\frac{1}{2}+\frac{1}{2}-(-\frac{1}{2})\}]$

$=\frac{3}{4}[2\sin {30}^o\cos {20}^o-\sin {70}^o]-\frac{3}{8}[\because \cos {20}^o=\sin {70}^o]$

$\frac{3}{4}.0-\frac{3}{8}=-\frac{3}{8}$