Question
If A = 30° and B = 60°, verify that.
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
✓
Answer
We have,
$\sin(\text{A}-\text{B})=\frac{1}{2}$ and $\cos(\text{A}+\text{B})=\frac{1}{2}$
Now, $\sin(\text{A}-\text{B})=\frac{1}{2}$
$\Rightarrow\sin(\text{A}-\text{B})=\sin30^\circ$
$\Rightarrow(\text{A}-\text{B})=30^\circ\dots(1)$
And, $\cos(\text{A}+\text{B})=\frac{1}{2}$
$\Rightarrow\cos(\text{A}+\text{B})=\cos60^\circ$
$\Rightarrow(\text{A}+\text{B})=60^\circ\dots(2)$
Adding (1) and (2), we get,
$\Rightarrow2\text{A}=90^\circ$
$\Rightarrow\text{A}=\frac{90^\circ}{2}=45^\circ$
Put $\text{A}=45^\circ$ in (2), we get,
$\Rightarrow45^\circ+\text{B}=60^\circ$
$\Rightarrow\text{B}=60^\circ-45^\circ$
$\Rightarrow\text{B}=15^\circ$
Thus, $\text{A}=45^\circ$ and $\text{B}=15^\circ$
$\sin(\text{A}-\text{B})=\frac{1}{2}$ and $\cos(\text{A}+\text{B})=\frac{1}{2}$
Now, $\sin(\text{A}-\text{B})=\frac{1}{2}$
$\Rightarrow\sin(\text{A}-\text{B})=\sin30^\circ$
$\Rightarrow(\text{A}-\text{B})=30^\circ\dots(1)$
And, $\cos(\text{A}+\text{B})=\frac{1}{2}$
$\Rightarrow\cos(\text{A}+\text{B})=\cos60^\circ$
$\Rightarrow(\text{A}+\text{B})=60^\circ\dots(2)$
Adding (1) and (2), we get,
$\Rightarrow2\text{A}=90^\circ$
$\Rightarrow\text{A}=\frac{90^\circ}{2}=45^\circ$
Put $\text{A}=45^\circ$ in (2), we get,
$\Rightarrow45^\circ+\text{B}=60^\circ$
$\Rightarrow\text{B}=60^\circ-45^\circ$
$\Rightarrow\text{B}=15^\circ$
Thus, $\text{A}=45^\circ$ and $\text{B}=15^\circ$
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