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 $	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

Similar Questions

If the angles of a triangle are in the ratio $4:1:1$,then the ratio of the longest side to the perimeter is

When current in a coil changes from $5\,A$ to $2\,A$ in $0.1\,s$,an average voltage of $50\,V$ is produced. The self-inductance of the coil is $............H$.

Suggest a possible explanation why the seeds in a pea pod are arranged in a row,whereas those in tomato are scattered in the juicy pulp.

The electrons which take part in order to exhibit variable oxidation states by transition metals are

Vedclass Test Series

Exam Paper Generator

Online Exam Module