The value of sin^{2} 51^{circ} + sin^{2} 39^{ | vedclass

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

Question

(A) We know that $\sin(90^{\circ} - 	heta) = \cos 	heta$. Given expression: $\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$. We can write $39^{\circ}$ a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We know that $\sin(90^{\circ} - 	heta) = \cos 	heta$. Given expression: $\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$. We can write $39^{\circ}$ as $(90^{\circ} - 51^{\circ})$. So,$\sin^{2} 39^{\circ} = \sin^{2}(90^{\circ} - 51^{\circ}) = \cos^{2} 51^{\circ}$. Substituting this into the expression: $\sin^{2} 51^{\circ} + \cos^{2} 51^{\circ}$. Using the identity $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get $1$.

The value of $\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$ is

Similar Questions

The values of $x$ in $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\sin x \cos x = \frac{1}{4}$ are $.......$

Which of the following is true?

Find the value of $\sin \frac{31 \pi}{3}$.

$\cos 18^{\circ} = $

Find the angle in radians through which a pendulum swings if its length is $75 \, cm$ and the tip describes an arc of length $21 \, cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module