Q62 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
$\cos ^3 110^{\circ}+\cos ^3 10^{\circ}+\cos ^3 130^{\circ}=$
- A$\frac{3}{4}$
- B$\frac{3}{8}$
- Correct answerC$\frac{3 \sqrt{3}}{8}$
- D$\frac{3 \sqrt{3}}{4}$
Solution
$\cos ^3 10+\cos ^3 110+\cos ^3 130^{\circ}$
We know that,
$$
\cos ^3 x+\cos ^3(120-x)+\cos ^3(120+x)=\frac{3}{4} \cos 3 x
$$
Here,
$$
x=10
$$
Now, $\cos ^3 10+\cos ^3 110^{\circ}+\cos ^3 130^{\circ}$
$$
\begin{aligned}
& =\cos ^3 10+\cos ^3\left(120^{\circ}-10^{\circ}\right) \cos ^3\left(120^{\circ}+10^{\circ}\right) \\
& =\left(\frac{3}{4}\right) \cos \left(3 \times 10^{\circ}\right)=\frac{3}{4} \cos 30^{\circ}=\frac{3}{4} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{8}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.