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Q.
If $\cos \, A = \frac{4}{5} , \cos \, B = \frac{12}{13}, \frac{3 \pi}{2} < A, B < 2 \pi , $ the values of the following : $\cos (A + B) $ is
Trigonometric Functions
Solution:
Since A and B both lie in the IV quadrant, it follows that sin A and sin B are negative.
Therefore,
$\sin A = -\sqrt{1-\cos^{2 }A}$
$ \Rightarrow \sin A = - \sqrt{1- \frac{16}{25}}$
$ = - \frac{3}{5}$
and $ \sin B = - \sqrt{1- \cos^{2} B}$
$ \Rightarrow \sin B = - \sqrt{1 - \frac{144}{169}} $
$= - \frac{5}{ 13} $
Now , $\cos\left(A + B\right)$
$ = \cos A \cos B - \sin A \sin B$
$ = \frac{4}{5} \times\frac{12}{13} - \left(\frac{-3}{5}\right)\left(\frac{-5}{13}\right)$
$ = \frac{33}{65} $