Question

(a) If Sin A = 12/13 and sin B = 4/5, where π/2 < A < π and 0 < B < π/2, find the following:
(i) sin (A + B) (ii) cos (A + B)

(b) If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).

Answer 1

(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