Question
Evaluate the following:
If A = 60° and B = 30°, verify that:
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$

Answer

$\text{A}=60^\circ$ and $\text{B}=30^\circ$
Now, $\text{A}+\text{B}=60^\circ+30^\circ=90^\circ$
Also, $\text{A}-\text{B}=60^\circ-30^\circ=30^\circ$
$\sin(\text{A}+\text{B})=\sin90^\circ=1$
$\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
$=\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
$=\Big(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac12\times\frac12\Big)=\Big(\frac34+\frac14\Big)=1$
$\therefore\ \sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$

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