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

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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 $\sqrt{2}$:$= \frac{(\sqrt{3} + 1) 	imes \sqrt{2}}{2\sqrt{2} 	imes \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{2 	imes 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

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

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