sin ^{2} 1^{circ} + sin ^{2} 3^{circ} + sin ^ | vedclass

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

Question

(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.Using the trigonometric ide

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.Using the trigonometric identity $\sin(90^{\circ} - 	heta) = \cos 	heta$,we can rewrite the terms:$\sin ^{2} 87^{\circ} = \sin ^{2}(90^{\circ} - 3^{\circ}) = \cos ^{2} 3^{\circ}$$\sin ^{2} 89^{\circ} = \sin ^{2}(90^{\circ} - 1^{\circ}) = \cos ^{2} 1^{\circ}$Substituting these into the original expression:$= \sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ} + \cos ^{2} 1^{\circ}$Grouping the terms with the same angle:$= (\sin ^{2} 1^{\circ} + \cos ^{2} 1^{\circ}) + (\sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ})$Using the identity $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$:$= 1 + 1 = 2$.

$\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ} = \ldots$

Similar Questions

$\cot \theta \cdot \tan \theta = \dots$

Write 'True' or 'False' and justify your answer.
The value of $\sin \theta + \cos \theta$ is always greater than $1$.

If $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$ and $0 < \theta < 90^{\circ}$,then the value of $\theta$ is ...... (in $^{\circ}$)

Given that $\alpha + \beta = 90^{\circ}$,show that $\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sin \alpha$.

Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$

Vedclass Test Series

Exam Paper Generator

Online Exam Module