find the following:
Question:

If $\cos A=-\frac{24}{25}$ and $\cos B=\frac{3}{5}$, where $\pi<A<\frac{3 \pi}{2}$ and $\frac{3 \pi}{2}<B<2 \pi$, find the following:

(i) $\sin (A+B)$

(ii) $\cos (A+B)$

Solution:

Given:

$\cos A=-\frac{24}{25} \quad$ and $\quad \cos B=\frac{3}{5}$

and $\pi<A<\frac{3 \pi}{2} \quad$ and $\quad \frac{3 \pi}{2}<B<2 \pi$.

That is, $A$ is in third quadrant and $\mathrm{B}$ is in fourth qudrant.

We know that $\sin e$ function is negative in third and fourth quadrant $s$.

Therefore,

$\sin A=-\sqrt{1-\cos ^{2} A} \quad$ and $\quad \sin B=-\sqrt{1-\cos ^{2} B}$

$\Rightarrow \sin A=\sqrt{1-\left(\frac{-24}{25}\right)^{2}} \quad$ and $\quad \sin B=-\sqrt{1-\left(\frac{3}{5}\right)^{2}}$

$\Rightarrow \sin A=-\sqrt{1-\frac{576}{625}} \quad$ and $\quad \sin B=-\sqrt{1-\frac{9}{25}}$

$\Rightarrow \sin A=-\sqrt{\frac{49}{625}} \quad$ and $\quad \sin B=-\sqrt{\frac{16}{25}}$

$\Rightarrow \sin A=\frac{-7}{25} \quad$ and $\quad \sin B=\frac{-4}{5}$

Now,

(i) $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$=\frac{-7}{25} \times \frac{3}{5}+\frac{-24}{25} \times \frac{-4}{5}$

$=\frac{-21}{125}+\frac{96}{125}$

$=\frac{75}{125}$

$=\frac{3}{5}$

(ii) $\cos (A+B)=\cos A \cos B-\sin A \sin B$

$=\frac{-24}{25} \times \frac{3}{5}-\frac{-7}{25} \times \frac{-4}{5}$

$=\frac{-72}{125}-\frac{28}{125}$

$=\frac{-100}{125}$

$=\frac{-4}{5}$

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