sin 60^{circ} sin 45^{circ} + cos 60^{circ} c | vedclass

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

Question

(C) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\cos 60^{\circ} = \frac{1}{2}$,an

Vedclass · Accepted Answer

$\frac{\sqrt{6} + \sqrt{2}}{4}$

Vedclass · Answer

(C) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\cos 60^{\circ} = \frac{1}{2}$,and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$.Substituting these values into the expression:$\sin 60^{\circ} \sin 45^{\circ} + \cos 60^{\circ} \cos 45^{\circ} = \left( \frac{\sqrt{3}}{2} ight) \left( \frac{1}{\sqrt{2}} ight) + \left( \frac{1}{2} ight) \left( \frac{1}{\sqrt{2}} ight)$$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$To rationalize the denominator,multiply the numerator and denominator by $\sqrt{2}$:$= \frac{(\sqrt{3} + 1) 	imes \sqrt{2}}{2\sqrt{2} 	imes \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{2 	imes 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

$\sin 60^{\circ} \sin 45^{\circ} + \cos 60^{\circ} \cos 45^{\circ} = \ldots \ldots \ldots \ldots$

Similar Questions

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

If $\tan \theta + \sec \theta = l$,then prove that $\sec \theta = \frac{l^{2} + 1}{2l}$.

If $\sin 70^{\circ} = \cos \theta$,then $\theta = \ldots \ldots \ldots \ldots$ (in $^{\circ}$)

Show that $\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta$

$(\sin 80^{\circ} + \cos 10^{\circ})(\sin 80^{\circ} - \cos 10^{\circ}) = \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module