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If $ A=30^{\circ} $ and $ B=60^{\circ} $, verify that$ \sin (A+B)=\sin A \cos B+\cos A \sin B $
Given:
\( A=30^{\circ} \) and \( B=60^{\circ} \)
To do:
We have to verify that \( \sin (A+B)=\sin A \cos B+\cos A \sin B \).
Solution:
We know that,
$\sin 30^{\circ}=\frac{1}{2}$
$\cos 30^{\circ}=\frac{\sqrt3}{2}$
$\sin 60^{\circ}=\frac{\sqrt3}{2}$
$\cos 60^{\circ}=\frac{1}{2}$
Let us consider LHS,
$\sin (A+B)=\sin (30^{\circ}+60^{\circ})$
$=\sin 90^{\circ}$
$=1$ (Since $\sin 90^{\circ}=1$)
Let us consider RHS,
$\sin A \cos B+\cos A \sin B=\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}$
$=\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) +\left(\frac{\sqrt3}{2}\right) \times \left(\frac{\sqrt{3}}{2}\right)$
$=\frac{1}{4} +\frac{3}{4}$
$=1$
LHS $=$ RHS
Hence proved.
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