Question

(a) If $sinA=\frac{12}{13}$ and $sinB=\frac{4}{5},$ where $\frac{\pi }{2}<A<\pi$ and $0<B<\frac{\pi }{2},$ find the following:
(i) sin(A+B)
(ii) cos(A+B)
(b) If sinA= $\frac{3}{5}$, $cosB=\frac{12}{13}$, where A and B both lie in second quadrant, find the value of sin(A+B).

Answer 1

(a) $sinA=\frac{12}{13}$ and $sinB=\frac{4}{5}$
$\therefore cosA=-\sqrt{1-si{n}^{2}A}$ and $cosB=-\sqrt{1-si{n}^{2}B}$
[ $\because$ In the second quadrant $cos\theta$ is negative ]
$\Rightarrow cosA=-\sqrt{1-{\left(\frac{12}{13}\right)}^2}$
$cosB=\sqrt{1-{\left(\frac{4}{5}\right)}^2}$
$\Rightarrow cosA=-\sqrt{1-\frac{144}{169}}$
$cosB=\sqrt{1-\frac{16}{15}}$
$\Rightarrow cosA=-\sqrt{\frac{25}{169}}$ and
$cosB=\sqrt{\frac{9}{25}}$
$\Rightarrow cosA=\frac{-5}{13}$ and $cosB=\frac{3}{5}$
Now,
$sin(A+B)=sinAcosB+cosAsinB$
$=\frac{12}{13}\times -\frac{5}{13}\times \frac{4}{5}$
$=\frac{36}{65}-\frac{20}{25}$
$=\frac{16}{65}$
(ii) $cos(A+B)=cosAcosB-sinAsinB$
$=\frac{-5}{13}\times \frac{3}{5}-\frac{12}{13}\times \frac{4}{5}$
$=\frac{-15}{65}-\frac{48}{65}$
$=\frac{-63}{65}$
(b) We have,
$sinA=\frac{3}{5}$ and $cosB=\frac{-12}{13}$
$\therefore cosA=-\sqrt{1-si{n}^{2}A}$ and $sinB=\sqrt{1-co{s}^{2}B}$
[ $\because$ In the second quadrant $cos\theta$ is negative]
$\Rightarrow cosA=-\sqrt{1-{\left(\frac{3}{5}\right)}^2}$
$sinB=\sqrt{1-{\left(\frac{-12}{13}\right)}^2}$
$\Rightarrow cosA=-\sqrt{1-\frac{9}{25}}$ and $sinB=\sqrt{1-\frac{144}{169}}$
$\Rightarrow cosA=-\sqrt{\frac{16}{25}}$ and
$sinB=\sqrt{\frac{25}{169}}$
$\Rightarrow cosA=-\frac{4}{5}$ and $sinB=\frac{5}{13}$
Now,
$sin(A+B)=sinAcosB+cosAsinB$
$\frac{3}{5}\times \left(\frac{-12}{13}\right)-\frac{4}{5}\times \frac{5}{13}$
$=-\frac{36}{65}-\frac{20}{65}=-\frac{56}{65}$
$\therefore sin(A+B)=-\frac{56}{65}$