if sin (A) = 5/13 with 90° < A < 180° and cos(B) = 5/13 with 270° < B < 360°?

Question

Find cos(A + B) 

Answer 1

sin A = 5/13 with 90⁰ < A < 180⁰  and cos(B) = 5/13 with 270⁰ < A < 360⁰ .

If sin A = 5/13, then cos A = √(1 - sin² A) = √(1 - (5/13)²) = 12/13.

⇒ cos A = - 12/13.                       (Since, the function of cos is negative in second quadrant)

If cos B = 5/13, then sin B = √(1 - cos² B) = √(1 - (5/13)²) = 12/13.

⇒ sin B = - 12/13.                       (Since, the function of sin is negative in fourth quadrant)

Using composite formula : cos (A + B) = cos A cos B - sin A sin B.

⇒ cos (A + B) = (- 12/13)(5/13) - (5/13)(- 12/13)

= - 60/169 + 60/169

= 0.

Therefore, cos(A + B) = 0.