Question

Diketahui sin a= 3/5 dan cos B = -8/7, sudut a sudut lancip dan sudut B sudut tumpul. hitunglah nilai sin (a+B) dan sin (a-B)

Answer 1

.
sin A  = 3/5 , A lancip
cos A = √(1- sin² A) = 4/5

cos B = - 8/17, B tumpul
sin B = √(1- cos² x) = 15/17

sin(A+B) = sin A cos B + cos A sin B
sin(A+B) = (3/5)(-8/17) + ( 4/5)(15/17) = 40/85

sin(A-B) = sin A cos B - cos A sin B
sin(A-B) = (3/5)(-8/17) - (4/5)(15/17) = - 84/85