We know that cos A = – √(1 – sin2 A) and cos B = √(1 – sin2 B)
So 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)
We know that sin (A + B) = sin A cos B + cos A sin B
So,
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)
We know that cos (A +B) = cos A cos B – sin A sin B
So,
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:
sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant.
We know that cos A = – √(1 – sin2 A) and sin B = √(1 – cos2 B)
So 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
We need to find sin (A + B)
Since, sin (A + B) = sin A cos B + cos A sin B
= 3/5 × (-12/13) + (-4/5) × 5/13
= -36/65 – 20/65
= -56/65