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Q.
If $\sin A =\frac{3}{5}, 0< A <\frac{\pi}{2}$ and $\cos B =\frac{-12}{13}, \pi< B <\frac{3 \pi}{2}$, then
value of $\sin ( A - B )$ is
Trigonometric Functions
Solution:
We have $: \sin A =\frac{3}{5}$, where $0< A <\frac{\pi}{2}$
$\therefore \cos A=\pm \sqrt{1-\sin ^{2} A} $
$\Rightarrow \cos A=+\sqrt{1-\sin ^{2} A}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$[\because \cos$ is positive in first quadrant $]$
It is given that $: \cos B=\frac{-12}{13}$ and $\pi< B< \frac{3 \pi}{2}$
$\therefore \sin B =\pm \sqrt{1-\cos ^{2} B } \Rightarrow \sin B =-\sqrt{1-\cos ^{2} B }$
$[\because$ Sine is negative in the third quadrant]
$\Rightarrow \sin B =-\sqrt{1-\left(\frac{-12}{13}\right)^{2}}=-\frac{5}{13}$
Now, $ \sin (A-B) =\sin A \cos B-\cos A \sin B$
$=\frac{3}{5} \times \frac{-12}{13}-\frac{4}{5} \times \frac{-5}{13}=-\frac{16}{65}$