Q49 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics
If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$ then
$\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2=$
- A$1$
- B$\frac{3}{4}$
- Correct answerC$\frac{9}{16}$
- D$\frac{9}{8}$
Solution
Let $x=\cos \alpha+i \sin \alpha, y=\cos \beta+i \sin \beta$ and
$z=\cos \gamma+i \sin \gamma \therefore x+y+z=0 \therefore x^3+y^3+z^3=3 x y z$
$\Rightarrow(\cos \alpha+i \sin \alpha)^3+(\cos \beta+i \sin \beta)^3+(\cos \gamma+i \sin \gamma)^3$
$=3(\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)(\cos \gamma+i \sin \gamma)$
$\Rightarrow(\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma)+i(\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma)$
$=3 \cos (\alpha+\beta+\gamma)+3 i \sin (\alpha+\beta+\gamma)$
$\therefore \cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$
$\Rightarrow 4 \cos ^3 \alpha-3 \cos \alpha+4 \cos ^3 \beta-3 \cos \beta+4 \cos ^3 \gamma-3 \cos \gamma$
$=3 \cos (\alpha+\beta+\gamma)$
$\Rightarrow 4\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)=3 \cos (\alpha+\beta+\gamma)$
$\Rightarrow \cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma=\frac{3}{4} \cos (\alpha+\beta+\gamma)$
Similarly, $\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma=\frac{-3}{4} \sin (\alpha+\beta+\gamma)$
Now, $\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2$
$=\frac{9}{16} \cos ^2(\alpha+\beta+\gamma)+\frac{9}{16} \sin ^2(\alpha+\beta+\gamma)=\frac{9}{16}$
Hence, the correct answer is option: 'C'.