The value of 18^ is

MCQ

The value of $\sin 18^{\circ}$ is

A
$\frac{\sqrt{5}+1}{4}$
✓
$\frac{\sqrt{5}-1}{4}$
C
$\frac{4}{\sqrt{5}+1}$
D
$\frac{4}{\sqrt{5}-1}$

✓

Answer

Correct option: B.

$\frac{\sqrt{5}-1}{4}$

(b) : Let $x=18^{\circ} \Rightarrow 5 x=5 \times 18^{\circ}=90^{\circ}$
$
\begin{aligned}
& \Rightarrow 2 x+3 x=90^{\circ} \Rightarrow 2 x=90^{\circ}-3 x \\
& \Rightarrow \sin 2 x=\sin \left(90^{\circ}-3 x\right) \Rightarrow \sin 2 x=\cos 3 x \\
& \Rightarrow 2 \sin x \cos x=4 \cos ^3 x-3 \cos x \\
& \Rightarrow 2 \sin x \cos x=\cos x\left(4 \cos ^2 x-3\right) \\
& \Rightarrow 2 \sin x=4 \cos ^2 x-3 \Rightarrow 2 \sin x=4\left(1-\sin ^2 x\right)-3 \\
& \Rightarrow 2 \sin x=4-4 \sin ^2 x-3 \Rightarrow 4 \sin ^2 x+2 \sin x-1=0
\end{aligned}
$
This is a quadratic in $\sin x$, where $a=4, b=2, c=-1$
$
\begin{aligned}
\therefore \quad \sin x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-2 \pm \sqrt{4-4(4)(-1)}}{8} \\
\quad=\frac{-2 \pm \sqrt{20}}{8}=\frac{-2 \pm 2 \sqrt{5}}{8}=\frac{-1 \pm \sqrt{5}}{4}
\end{aligned}
$
But $\sin 18^{\circ}>0$ because $18^{\circ}$ lies in first quadrant.
$
\therefore \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}
$

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