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Question
If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)
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Solution
Given:
\[\sin A = \frac{12}{13}\text{ and }\sin B = \frac{4}{5}\]
\[\text{When, }\frac{\pi}{2} < A < \pi\text{ and }0 < B < \frac{\pi}{2}, \]
\[\cos A = - \sqrt{1 - \sin^2 A}\text{ and }\cos B = \sqrt{1 - \sin^2 B}\]
( As cosine function is negative in second qudrant and positive in first quadrant )
\[\Rightarrow \cos A = - \sqrt{1 - \left( \frac{12}{13} \right)^2}\text{ and }\cos B = \sqrt{1 - \left( \frac{4}{5} \right)^2}\]
\[ \Rightarrow \cos A = - \sqrt{1 - \frac{144}{169}}\text{ and }\cos B = \sqrt{1 - \frac{16}{25}}\]
\[ \Rightarrow \cos A = - \sqrt{\frac{25}{169}}\text{ and }\cos B = \sqrt{\frac{9}{25}}\]
\[ \Rightarrow \cos A = \frac{- 5}{13}\text{ and }\cos B = \frac{3}{5}\]
Now,
\[\sin\left( A + B \right) = \sin A \cos B + \cos A \sin B\]
\[ = \frac{12}{13} \times \frac{3}{5} + \frac{- 5}{13} \times \frac{4}{5}\]
\[ = \frac{36}{65} + \frac{- 20}{65}\]
\[ = \frac{16}{65}\]
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