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

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Question

(B) Let $A = 18^{\circ}$. Then $5A = 90^{\circ}$,which implies $2A = 90^{\circ} - 3A$. Taking sine on both sides,$\sin 2A = \sin(90^{\circ} - 3A) = \c

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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