sin 75^circ = | vedclass

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Question

(B) To find the value of $\sin 75^\circ$,we can express $75^\circ$ as the sum of two standard angles: $45^\circ + 30^\circ$.Using the trigonometric id

Vedclass · Accepted Answer

$\frac{\sqrt{3} + 1}{2\sqrt{2}}$

Vedclass · Answer

(B) To find the value of $\sin 75^\circ$,we can express $75^\circ$ as the sum of two standard angles: $45^\circ + 30^\circ$.Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:$\sin 75^\circ = \sin(45^\circ + 30^\circ)$$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$$= (\frac{1}{\sqrt{2}} 	imes \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} 	imes \frac{1}{2})$$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$.

$\sin 75^\circ = $

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