Question

if sinA = 3/5 and cosB = 9/41, 0<A<π/2,and -π/2<B<0 then sin(A-B)is equal to
1) 84/205
2) 94/205
3) 187/205
4) 181/205​

Answer 1

Step-by-step explanation:

$\sin(a) = \frac{3}{5}$

${ \sin }^{2} a + { \cos }^{2} a = 1$

${ \cos }^{2} a = 1 - { \frac{3}{5} }^{2}$

$= 1 - \frac{9}{25}$

$= 16 \div 25$

$\cos(a) = \frac{4}{5}$

$\cos(b ) = \frac{9}{41}$

${ \sin }^{2} b + { \cos }^{2} b = 1$

${ \sin }^{2} b = 1 - ( { \frac{9}{41}) }^{2}$

$= 1 - \frac{81}{1631}$

$= \frac{1600}{1681}$

$\sin(b) = \frac{40}{41}$

$\sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b)$

$= \frac{3}{5} \times \frac{9}{41} - \frac{4}{5} \times \frac{40}{41}$

$\frac{27 - 160}{205}$

$= \frac{ - 133}{205}$