The value of sin 18^{circ} is | vedclass

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(B) Let $	heta = 18^{\circ}$.Then $2	heta = 36^{\circ}$ and $3	heta = 54^{\circ}$.Since $2	heta + 3	heta = 90^{\circ}$,we have $2	heta = 90^{\ci

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $	heta = 18^{\circ}$.Then $2	heta = 36^{\circ}$ and $3	heta = 54^{\circ}$.Since $2	heta + 3	heta = 90^{\circ}$,we have $2	heta = 90^{\circ} - 3	heta$.Taking sine on both sides: $\sin 2	heta = \sin(90^{\circ} - 3	heta) = \cos 3	heta$.Using the double angle and triple angle formulas: $2\sin	heta\cos	heta = 4\cos^3	heta - 3\cos	heta$.Since $\cos 18^{\circ} 
eq 0$,we can divide by $\cos	heta$: $2\sin	heta = 4\cos^2	heta - 3$.Substitute $\cos^2	heta = 1 - \sin^2	heta$: $2\sin	heta = 4(1 - \sin^2	heta) - 3$.$2\sin	heta = 4 - 4\sin^2	heta - 3 \Rightarrow 4\sin^2	heta + 2\sin	heta - 1 = 0$.Using the quadratic formula for $\sin	heta$: $\sin	heta = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.Since $18^{\circ}$ is in the first quadrant,$\sin 18^{\circ} > 0$,so we take the positive root.Therefore,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.

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

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