Calculus Question

Question

Use the given data to find the values of the cosine of (A + B) and of (A – B). Give the quadrant in which each of the angles (A + B) and (A – B) terminates. Angle A in Quadrant II, tan A = -12/5 , Angle B in Quadrant III, sin B = - 3/5

1. cos (A + B) =

a. 56/65

b. -16/65

c. -56/65

d. 16/65

2.cos (A – B) =

a. -16/65

b. -56/65

c. 56/65

d. 16/65

My answer for question 1 is a, and question 2 is a.

Answer 1

For Angle A in Quadrant II with tan (-12/5), a vector to (-12,5) would make an angle of (90 + 90 − 67.38013505)° or 112.619865° (Quadrant II) with the positive x-axis.

For Angle B in Quadrant III with sin (-3/5), a vector to (-4,-3) would make an angle of (90 + 90 + 36.86989765)° or 216.8698977° (Quadrant III) with the positive x-axis. [Note that 5² − (-3)² gives (-4)² for coordinates (-4,-3).]

Next, cos (A + B) gives cos ( 112.619865° + 216.8698977°) equal to 0.8615384625 or 56/65 (Choice A).

Finally, cos (A − B) gives cos ( 112.619865° − 216.8698977°) equal to -0.2461538462 or -16/65 (Choice A).

Answer 2

Hi,

tan A = -12/5 = opp /adj ----> opp = 12 adj = 5 and Hyp = (12^2 + 5^2 ) ^ 1/2 = 13

now, you can find cos A = -5/13 and sin A = 12/13 because A is in II quadrant

sin B = -3/5 = opp/hyp ----> opp = 3 and hyp = 5 and hyp^2 = opp^2 + adj^2

25 = 9 + adj^2

16 = adj^2 ----> adj = 4

now you can find cos B = - 4 / 5 because B is in III quadrant

cos ( A + B ) = cos A cos B - sin A sin B = -5/13 * -4/5 - 12/13 * -3/5 = 56/65 choice a is right

cos ( A - B ) = cos A cos B + sin A sin B = -5/13 * -4/5 + 12/13 * -3/5 = -16/65 choice a is right

I hope it is useful,

Minoo