Question

Sin A=15/17 and cos B=-4/5 then find sin(A+B),cos(A-B) and tan (A+B)

Answer 1

$\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b) \\ = \frac{15}{17} \times \frac{ - 4}{5} + \frac{8}{17} \times \frac{3}{5} \ = \ \frac{ - 12}{17} + \frac{24}{85} \\ = \frac{ - 60 + 24}{85} \\ = \frac{ - 36}{85}$

$\cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b) \\ = \frac{8}{17} \times \frac{ - 4}{5} + \frac{15}{17} \times \frac{3}{5} \\ = \frac{ - 32}{85} + \frac{45}{85} \\ = \frac{13}{85}$

$\tan(a + b) = \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) }$

$\tan(a) = \frac{ \sin(a) }{ \cos(a) } = \frac{15}{8} \\ \\ \tan(b) = \frac{ \sin(b) }{ \cos(b) } \\ = \frac{ - 3}{4}$

therefore

$\tan(a + b) = \frac{ \frac{15}{8} + \frac{ - 3}{4} }{1 - \frac{15}{8} \times \frac{ - 3}{4} } \\ \frac{ \frac{11}{8} }{1 + \frac{45}{32} } \\ = \frac{ \frac{11}{8} }{ \frac{77}{32} } \\ = \frac{11}{8} \times \frac{32}{77} \\ = \frac{4}{7}$