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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

Evaluate : $\sin 18^{\large\circ}$

$\begin{array}{1 1}(A)\;\large\frac{\sqrt 5-1}{4}\\(B)\;\large\frac{\sqrt 5+1}{4}\\(C)\;\large\frac{\sqrt 7-1}{4}\\(D)\;\large\frac{\sqrt 7+1}{4}\end{array} $

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A)
Let $\theta=18^{\large\circ}$ then $5\theta=90^{\large\circ}$
$2\theta+3\theta=90^{\large\circ}$
$2\theta=90^{\large\circ}-3\theta$
$\sin 2\theta=\sin (90^{\large\circ}-3\theta)$
$\sin 2\theta=\cos3\theta$
$2\sin \theta\cos \theta=4\cos^3\theta-3\cos\theta$
Dividing by $\cos\theta$ both side
$2\sin \theta=4\cos^2\theta-3$
$2\sin \theta=4(1-\sin ^2\theta)-3$
$2\sin \theta=1-4\sin ^2\theta$
$4\sin^2\theta+2\sin \theta-1=0$
$\therefore \sin \theta=\large\frac{-2\pm \sqrt{4+16}}{8}=\frac{-2\pm 2\sqrt 5}{8}$
$\Rightarrow \large\frac{-1\pm \sqrt 5}{4}$
$\sin \theta=\large\frac{-1+\sqrt 5}{4}$ or $\large\frac{-1-\sqrt 5}{4}$
$\theta=18^{\large\circ}$
$\therefore \sin \theta=\sin 18^{\large\circ} > 0$ for $18^{\large\circ}$ in the first quadrant
$\therefore \sin \theta$ i.e $\sin 18^{\large\circ}=\large\frac{\sqrt 5-1}{4}$
Hence (A) is the correct answer.
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