sin ^{2} 17.5^{circ} + sin ^{2} 72.5^{circ} i | vedclass

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

Question

(B) We have,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ Since $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can write $\sin(72.5^{\circ}) = \sin(

Vedclass · Accepted Answer

$	an ^{2} 45^{\circ}$

Vedclass · Answer

(B) We have,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ Since $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can write $\sin(72.5^{\circ}) = \sin(90^{\circ} - 17.5^{\circ}) = \cos(17.5^{\circ})$. Therefore,$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ} = \sin ^{2} 17.5^{\circ} + \cos ^{2} 17.5^{\circ}$. Using the identity $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$,we get $1$. Since $	an 45^{\circ} = 1$,then $	an ^{2} 45^{\circ} = 1^{2} = 1$. Thus,the expression is equal to $	an ^{2} 45^{\circ}$.

$\sin ^{2} 17.5^{\circ} + \sin ^{2} 72.5^{\circ}$ is equal to

Similar Questions

The value of $\sin 18^{\circ}$ is

Find the degree measure corresponding to the following radian measure (Use $\pi = \frac{22}{7}$):
$\frac{5 \pi}{3}$ (in $^{\circ}$)

If $\sin A = -\frac{60}{61}$,$\cot B = -\frac{40}{9}$ and neither $A$ nor $B$ is in the $4^{\text{th}}$ quadrant,then $6 \cot A + 4 \sec B = $

What are the polar coordinates of $(-\sqrt{3}, -1)$?

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