Question
If A = 30° and B = 60°, verify that.
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
✓
Answer
Given
$\text{A}=30^\circ$ and $\text{B}=60^\circ\dots(1)$
To verify:
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\dots(2)$
Now consider LHS of the expression to be verified in equation (2)
Therefore,
$\sin(\text{A}+\text{B})=\sin(30+60)$
$=\sin90$
$=1$
Now consider RHS of the expression to be verified in equation (2)
Therefore
$\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}=\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ $
$=\frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$
$=\frac{1+3}{4}$
$=1$
Hence it is verified that,
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
$\text{A}=30^\circ$ and $\text{B}=60^\circ\dots(1)$
To verify:
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\dots(2)$
Now consider LHS of the expression to be verified in equation (2)
Therefore,
$\sin(\text{A}+\text{B})=\sin(30+60)$
$=\sin90$
$=1$
Now consider RHS of the expression to be verified in equation (2)
Therefore
$\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}=\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ $
$=\frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$
$=\frac{1+3}{4}$
$=1$
Hence it is verified that,
$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
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