Question

If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (A + B)

Answer 1

Given:
\[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\]
\[\text{ and }\pi < A < \frac{3\pi}{2}\text{ and }\frac{3\pi}{2} < B < 2\pi . \]
That is, A is in third quadrant and B is in fourth qudrant. 
We know that sine function is negative in third and fourth quadrants. 
Therefore, 
\[\sin A = - \sqrt{1 - \cos^2 A}\text{ and }\sin B = - \sqrt{1 - \cos^2 B}\]
\[ \Rightarrow \sin A = \sqrt{1 - \left( \frac{- 24}{25} \right)^2}\text{ and }\sin B = - \sqrt{1 - \left( \frac{3}{5} \right)^2}\]
\[ \Rightarrow \sin A = - \sqrt{1 - \frac{576}{625}}\text{ and }\sin B = - \sqrt{1 - \frac{9}{25}}\]
\[ \Rightarrow \sin A = - \sqrt{\frac{49}{625}}\text{ and }\sin B = - \sqrt{\frac{16}{25}}\]
\[ \Rightarrow \sin A = \frac{- 7}{25}\text{ and }\sin B = \frac{- 4}{5}\]
Now
\[ \sin\left( A + B \right) = \sin A \cos B + \cos A \sin B\]
\[ = \frac{- 7}{25} \times \frac{3}{5} + \frac{- 24}{25} \times \frac{- 4}{5}\]
\[ = \frac{- 21}{125} + \frac{96}{125}\]
\[ = \frac{75}{125}\]
\[\frac{3}{5}\]