Question

If $cosA=-\frac{24}{25}$ and $cosB=\frac{3}{5}$, 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)

Answer 1

We have,
$cosA=-\frac{24}{25}$ and $cosB=\frac{3}{5}$
$\therefore sinA=-\sqrt{1-co{s}^{2}A}$
$sinB=-\sqrt{1-co{s}^{2}B}$
[$\because$ In the 3rd and 4th quadrant $sin\theta$ is negative]
$\Rightarrow sinA=-\sqrt{1-{(-\frac{24}{25})}^2}$ and
$sinB=-\sqrt{1-{\left(\frac{3}{5}\right)}^2}$
$\Rightarrow sinA=-\sqrt{1-\frac{576}{625}}$ and
$sinB=-\sqrt{\frac{9}{25}}$
$\Rightarrow sinA=-\sqrt{\frac{49}{625}}$ and
$sinB=-\sqrt{\frac{16}{25}}$
$\Rightarrow sinA=-\frac{7}{25}$ and $sinB=-\frac{4}{5}$
Now,
(i) $sin(A+B)=sinAcosB+cosAsinB$
$=-\frac{7}{25}\times \frac{3}{5}-\frac{24}{25}-(-\frac{4}{5})$
$=-\frac{21}{25}+\frac{96}{125}$
$=\frac{75}{125}=\frac{3}{5}$
(ii) $cos(A+B)=cosAcosB-sinAsinB$
$=-\frac{24}{25}\times \frac{3}{5}-(-\frac{7}{25})\times (-\frac{4}{5})$
$=-\frac{72}{125}-\frac{28}{125}$
$=\frac{-72-28}{125}=\frac{-100}{125}=\frac{-4}{5}$