Question

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

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

Answer 1

The correct option is D $\frac{-3}{8}$

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

$=(\sin {10}^{\circ }+\sin {50}^{\circ })[{(\sin {10}^{\circ }+\sin {50}^{\circ })}^2-3\sin {10}^{\circ }\sin {50}^{\circ }]-{\sin }^3{70}^{\circ }$ using $a^3+b^3=(a+b)[{(a+b)}^2-3ab]$

$=2\sin {30}^{\circ }\cos {20}^{\circ }[{\left(2\sin {30}^{\circ }\cos {20}^{\circ }\right)}^2-3\sin {10}^{\circ }\sin {50}^{\circ }]-{\sin }^3{70}^{\circ }$

$=2\times \frac{1}{2}\cos {20}^{\circ }[{(2\times \frac{1}{2}\cos {20}^{\circ })}^2-3\sin {10}^{\circ }\sin {50}^{\circ }]-{\sin }^3{70}^{\circ }$

$=\cos {20}^{\circ }[{\left(\cos {20}^{\circ }\right)}^2-3\sin {10}^{\circ }\sin {50}^{\circ }]-{\sin }^3{70}^{\circ }$

$={\cos }^3{20}^{\circ }-3\sin {10}^{\circ }\sin {50}^{\circ }\cos {20}^{\circ }-{\sin }^3{70}^{\circ }$

$={\cos }^3({90}^{\circ }-{70}^{\circ })-3\sin {10}^{\circ }\sin {50}^{\circ }\cos {20}^{\circ }-{\sin }^3{70}^{\circ }$

$={\sin }^3{70}^{\circ }-3\sin {10}^{\circ }\sin {50}^{\circ }\cos {20}^{\circ }-{\sin }^3{70}^{\circ }$

$=-3\sin {10}^{\circ }\sin {50}^{\circ }\cos {20}^{\circ }$

$=-3\sin {10}^{\circ }\sin {50}^{\circ }\cos ({90}^{\circ }-{70}^{\circ })$

$=-3\sin {10}^{\circ }\sin {50}^{\circ }\sin {70}^{\circ }$

$=\frac{-3\sin {80}^{\circ }}{8\sin {10}^{\circ }}$

$=\frac{-3\sin {80}^{\circ }}{8\sin ({90}^{\circ }-{80}^{\circ })}$

$=\frac{-3\sin {80}^{\circ }}{8\sin {80}^{\circ }}=\frac{-3}{8}$