Evaluate:sin 25^{circ} cos 65^{circ} + cos 25 | vedclass

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

Question

(D) The given expression is $\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$.We know that $\cos(90^{\circ} - 	heta) = \sin 	heta$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The given expression is $\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$.We know that $\cos(90^{\circ} - 	heta) = \sin 	heta$ and $\sin(90^{\circ} - 	heta) = \cos 	heta$.Substituting $\cos 65^{\circ} = \cos(90^{\circ} - 25^{\circ}) = \sin 25^{\circ}$ and $\sin 65^{\circ} = \sin(90^{\circ} - 25^{\circ}) = \cos 25^{\circ}$:$= (\sin 25^{\circ})(\sin 25^{\circ}) + (\cos 25^{\circ})(\cos 25^{\circ})$$= \sin^{2} 25^{\circ} + \cos^{2} 25^{\circ}$Using the identity $\sin^{2} A + \cos^{2} A = 1$,we get:$= 1$

Evaluate:
$\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$

Similar Questions

$\sin 2A = 2 \sin A$ is true when $A =$ (in $^{\circ}$)

State whether the following is true or false. Justify your answer.
$\sin (A+B) = \sin A + \sin B$

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

Express $\cot 85^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$.

Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Evaluate:$\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$