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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Prove that $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$.

Find the radian measure corresponding to the following degree measure: $-47^{\circ} 30^{\prime}$.

If $\sqrt{3} \cos \theta + \sin \theta > 0$,then:

$\cos 36^{\circ} - \cos 72^{\circ}$ is equal to

If $\tan \theta = 2$ and $\theta$ lies in the third quadrant,then the value of $\sec \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module