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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II3 Marks
Question
$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$
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Answer

Let $\theta=18^{\circ}$
$\begin{aligned}
& \therefore 5 \theta=90^{\circ} \\
& \therefore 2 \theta+3 \theta=90^{\circ} \\
& \therefore 2 \theta=90^{\circ}-3 \theta \\
& \therefore \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right) \\
& \therefore \sin 2 \theta=\cos 3 \theta \\
& \therefore 2 \sin \theta \cos \theta=4 \cos ^3 \theta-3 \cos \theta \\
& \left.\therefore 2 \sin \theta=4 \cos ^2 \theta-3 \ldots . \because \cos \theta \neq 0\right] \\
& \therefore 2 \sin \theta=4\left(1-\sin ^2 \theta\right)-3 \\
& \therefore 2 \sin \theta=1-4 \sin ^2 \theta \\
& \therefore 4 \sin 2 \theta+2 \sin \theta-1=0 \\
& \therefore \sin \theta=\frac{-2 \pm \sqrt{4+16}}{8} \\
& =\frac{-2 \pm 2 \sqrt{5}}{8} \\
& =\frac{-1 \pm \sqrt{5}}{4}
\end{aligned}$
Since, $\sin 18^{\circ}>0$
$\therefore \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

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