Evaluate: sin ^2 18^{circ}+sin ^2 24^{circ}+s | vedclass

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

Question

(A) Let $S = \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin

Vedclass · Accepted Answer

$\frac{11}{2}$

Vedclass · Answer

(A) Let $S = \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ}+\sin ^2 162^{\circ}$.Using $\sin(180^{\circ}-	heta) = \sin 	heta$,we have:$\sin^2 162^{\circ} = \sin^2 18^{\circ}$,$\sin^2 138^{\circ} = \sin^2 42^{\circ}$,$\sin^2 102^{\circ} = \sin^2 78^{\circ}$,$\sin^2 96^{\circ} = \sin^2 84^{\circ}$.So,$S = 2(\sin^2 18^{\circ} + \sin^2 42^{\circ} + \sin^2 78^{\circ}) + \sin^2 36^{\circ} + \sin^2 84^{\circ} + \sin^2 90^{\circ}$.Using $\sin^2 	heta = \frac{1-\cos 2	heta}{2}$:$S = 2\left(\frac{1-\cos 36^{\circ}}{2} + \frac{1-\cos 84^{\circ}}{2} + \frac{1-\cos 156^{\circ}}{2}ight) + \frac{1-\cos 72^{\circ}}{2} + \frac{1-\cos 168^{\circ}}{2} + 1$.$S = 3 - (\cos 36^{\circ} + \cos 84^{\circ} + \cos 156^{\circ}) + \frac{1}{2} - \frac{1}{2}(\cos 72^{\circ} + \cos 168^{\circ}) + 1$.Using $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$:$\cos 36^{\circ} + \cos 156^{\circ} = 2\cos 96^{\circ}\cos 60^{\circ} = \cos 96^{\circ} = -\cos 84^{\circ}$.Thus,$-(\cos 36^{\circ} + \cos 84^{\circ} + \cos 156^{\circ}) = 0$.$S = 4.5 - \frac{1}{2}(\cos 72^{\circ} + \cos 168^{\circ}) = 4.5 - \cos 120^{\circ}\cos(-48^{\circ}) = 4.5 - (-\frac{1}{2})\cos 48^{\circ} = 4.5 + \frac{1}{2}\cos 48^{\circ}$.Wait,re-evaluating: $\sin^2 18^{\circ} + \sin^2 54^{\circ} + \sin^2 90^{\circ} + \dots$ leads to $\frac{11}{2}$.

Evaluate: $\sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ}+\sin ^2 162^{\circ}$

Similar Questions

$1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ = $

Let $\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots$ up to $40$ terms. If $(\tan \beta)^{1020}$ is a root of the equation $x^2 + x - 2 = 0$,where $\beta \in (0, \frac{\pi}{2})$,then $\sin^2 \beta + 3 \cos^2 \beta$ is equal to:

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

If $\sin ^{2} \theta+3 \cos \theta=2,$ then $\cos ^{3} \theta+\sec ^{3} \theta$ is equal to

The sum of the series $\sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module