Question

If sin A =
13 and sin
in B = }į
5, where < A < i and
0<B< , find the values of the following-
(i) sin (A + B)
(ii) cos (A + B)​

Answer 1

Answer:

(a) Given as Sin A = 12/13 and sin B = 4/5, where π/2 < A < π and 0 < B < π/2 As we know that cos A = – √(1 – sin2 A) and cos B = √(1 – sin2 B) Therefore let us find the value of cos A and cos B cos A = – √(1 – sin2 A) = – √(1 – (12/13)2) = – √(1 - 144/169) = -√((169 - 144)/169) = – √(25/169) = – 5/13 cos B = √(1 – sin2 B) = √(1 – (4/5)2) = √(1 - 16/25) = √((25 - 16)/25) =√(9/25) = 3/5 (i) sin (A + B) As we know that sin (A + B) = sin A cos B + cos A sin B Therefore, sin (A + B) = sin A cos B + cos A sin B = 12/13 × 3/5 + (-5/13) × 4/5 = 36/65 – 20/65 = 16/65 (ii) cos (A + B) As we know that cos (A + B) = cos A cos B – sin A sin B Therefore, cos (A + B) = cos A cos B – sin A sin B = -5/13 × 3/5 – 12/13 × 4/5 = -15/65 – 48/65 = – 63/65 (b) Given as sin A = 3/5, cos B = –12/13, here A and B, both lie in second quadrant. As we know that cos A = – √(1 – sin2 A) and sin B = √(1 – cos2 B) Therefore let us find the value of cos A and sin B cos A = – √(1 – sin2 A) = – √(1 – (3/5)2) = – √(1 - 9/25) = – √((25 - 9)/25) = – √(16/25) = – 4/5 sin B = √(1 – cos2 B) = √(1 – (-12/13)2) = √(1 – 144/169) = √((169 - 144)/169) = √(25/169) = 5/13 Now, we need to find sin (A + B) Since, sin (A + B) = sin A cos B + cos A sin B by formula = 3/5 × (-12/13) + (-4/5) × 5/13 = -36/65 – 20/65 = -56/65

Explanation:

idk if its wrong or right.