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

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

Question

(D) The given expression is $S = \sin^2 5^\circ + \sin^2 10^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$.There are $18$ terms in the series from

Vedclass · Accepted Answer

$9\frac{1}{2}$

Vedclass · Answer

(D) The given expression is $S = \sin^2 5^\circ + \sin^2 10^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$.There are $18$ terms in the series from $5^\circ$ to $90^\circ$ in steps of $5^\circ$.We know that $\sin^2 	heta + \sin^2(90^\circ - 	heta) = \sin^2 	heta + \cos^2 	heta = 1$.Pairing the terms: $(\sin^2 5^\circ + \sin^2 85^\circ) + (\sin^2 10^\circ + \sin^2 80^\circ) + \dots + (\sin^2 40^\circ + \sin^2 50^\circ) + \sin^2 45^\circ + \sin^2 90^\circ$.There are $8$ such pairs,each equal to $1$.So,the sum is $8(1) + \sin^2 45^\circ + \sin^2 90^\circ$.Substituting values: $8 + (\frac{1}{\sqrt{2}})^2 + (1)^2 = 8 + \frac{1}{2} + 1 = 9\frac{1}{2}$.

The value of $\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$ is equal to

Similar Questions

If $\sin A, \cos A$ and $\tan A$ are in $G.P.$,then $\cos^3 A + \cos^2 A$ is equal to

If $\frac{\cos x}{\cos (x-2 y)}=\lambda$,then $\tan (x-y) \tan y$ is equal to

$\frac{2\sin \theta \tan \theta (1 - \tan \theta ) + 2\sin \theta \sec^2 \theta}{(1 + \tan \theta)^2} = $

If $A = \{x \in [0, 2\pi] : \tan x - \tan^2 x > 0\}$ and $B = \{x \in [0, 2\pi] : |\sin x| < \frac{1}{2}\}$,then $A \cap B =$

The value of $36(4 \cos^2 9^{\circ}-1)(4 \cos^2 27^{\circ}-1)(4 \cos^2 81^{\circ}-1)(4 \cos^2 243^{\circ}-1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module