Q61 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
$\begin{aligned}
& \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+ \\
& \sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ}+\sin ^2 162^{\circ}=
\end{aligned}$
- Correct answerA$\frac{11}{2}$
- B$\frac{9}{2}$
- C5
- D4
Solution
$\begin{aligned} & \begin{array}{r}\text { Since, } \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ} \\ +\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2(180-78)^{\circ}+\sin ^2(180-42)^{\circ}+ \\ \sin ^2(180-18)^{\circ} \\ =1+2 \sin ^2 18^{\circ}+\sin ^2 36^{\circ}+\sin ^2 24^{\circ}+\sin ^2 96^{\circ} \\ +2 \sin ^2 42^{\circ}+2 \sin ^2 78^{\circ}\end{array} \\ & +\frac{1-\cos 192^{\circ}}{2}+1-\cos 84^{\circ}+1-\cos 15^{\circ} \\ & =1+2\left(\frac{\sqrt{5}-1}{4}\right)^2+\frac{10-2 \sqrt{5}}{16}+\frac{1-\cos 48^{\circ}}{2} \\ & \left(\cos 84^{\circ}-\cos 24^{\circ}\right) \\ & =3+\frac{2(5+1-2 \sqrt{5})}{16}+\frac{5-\sqrt{5}}{8}+\frac{1}{2}-\frac{1}{2} \cos 48^{\circ} \\ & -\frac{1}{2} \cos \left(180^{\circ}+12^{\circ}\right)+\frac{1}{2}-\cos 84^{\circ}-\cos \left(180^{\circ}-24^{\circ}\right) \\ & =4+\frac{6-2 \sqrt{5}}{8}+\frac{5 \sqrt{5}}{8}-\frac{1}{2}\left(\cos 48^{\circ}-\cos 12^{\circ}\right)- \\ & \quad\left(-2 \sin 54^{\circ} \cdot \sin 30^{\circ}\right) \\ & =4+\frac{11-3 \sqrt{5}}{8}-\frac{1}{2}\left(-2 \sin 30^{\circ} \cdot \sin 18^{\circ}\right)- \\ & =4+\frac{11-3 \sqrt{5}}{8}+\frac{\sqrt{5}-1}{8}+\frac{\sqrt{5}+1}{4}=4+\frac{3}{2}=\frac{11}{2} .\end{aligned}$
Hence, the correct answer is option: 'A'.