Precalculus Question

Question

sin(a)=12/13, and a is in quadrant II. tan(b)=3/4, and b is in quadrant III. What is cos(a+b)?

steps too please, because i this is new to me

Answer 1

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

sin(a) = 12/13 = opposite/hypotenuse. It's in Quadrant II, where cos is negative. Use Pythagorean Theorem or Pythagorean Triples to find the side adjacent to the angle is 5.

cos(a) = - 5/13

tan(b) is in Quadrant III, where sin and cos are negative. tan(b) = 3/4 = opposite/adjacent. Hypotenuse is 5.

sin(b) = - 3/5

cos(b) = - 4/5

cos(a+b) = (-5/13)((-4/5) - (12/13)(-3/5) =

4/13 - (-36/65) = 4/13 + 36/65 = 20/65 + 36/65 = 56/65