The value of sin^2 5^{circ} + sin^2 10^{circ} | vedclass

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

Question

(A) The given expression is $S = \sin^2 5^{\circ} + \sin^2 10^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ}$.We know that $\sin^2 	heta +

Vedclass · Accepted Answer

$\frac{19}{2}$

Vedclass · Answer

(A) The given expression is $S = \sin^2 5^{\circ} + \sin^2 10^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ}$.We know that $\sin^2 	heta + \sin^2(90^{\circ} - 	heta) = \sin^2 	heta + \cos^2 	heta = 1$.There are $17$ terms from $5^{\circ}$ to $85^{\circ}$ in steps of $5^{\circ}$.These can be paired as $(\sin^2 5^{\circ} + \sin^2 85^{\circ}) + (\sin^2 10^{\circ} + \sin^2 80^{\circ}) + \ldots + (\sin^2 40^{\circ} + \sin^2 50^{\circ}) + \sin^2 45^{\circ}$.There are $8$ such pairs,each summing to $1$,and one middle term $\sin^2 45^{\circ} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.So,the sum of these $17$ terms is $8 	imes 1 + \frac{1}{2} = 8.5$.Adding the last term $\sin^2 90^{\circ} = 1$,the total sum is $8.5 + 1 = 9.5 = \frac{19}{2}$.

The value of $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ} = $

Similar Questions

If $\sec \theta + \tan \theta = \frac{1}{3}$,then the quadrant in which $2 \theta$ lies is

If $3 \sin \theta + 4 \cos \theta = 3$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then $\sin 2 \theta = $

The number of solutions of the equation $2 \cos(e^x) = 5^x + 5^{-x}$ is:

If $\sin \theta = \frac{2t}{1+t^2}$ and $\theta$ lies in the second quadrant,then $\cos \theta$ is equal to

If $\sin x - \sin y = \frac{27}{65}$ and $\cos x - \cos y = \frac{-21}{65}$,then find the value of $\sin(x + y)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module