given-that-sinA-12-13-and-sinB-4-5-where-A-and-B-are-acute-angles-find-cos-A-B-and-sin-A-B-

Question Number 10837 by okhema last updated on 27/Feb/17

g i v e n t h a t s i n A = \frac{12}{13} a n d s i n B = \frac{4}{5},

w h e r e A a n d B a r e a c u t e a n g l e s,

f i n d c o s (A - B) a n d s i n (A + B)

Answered by ridwan balatif last updated on 27/Feb/17

sinA = 12 / 13 sinB = 4 / 5

cosA = 5 / 13 cosB = 3 / 5

\cos (A - B) = cosA \times cosB + sinA \times sinB

= (\frac{5}{13}) \times (\frac{3}{5}) + (\frac{12}{13}) \times (\frac{4}{5})

= \frac{15}{65} + \frac{48}{65} = \frac{63}{65}

\sin (A + B) = sinA \times cosB - sinB \times cosA

= \frac{12}{13} \times \frac{3}{5} - \frac{4}{5} \times \frac{5}{13}

= \frac{36}{65} - \frac{20}{65} = \frac{16}{65}