Please Help! Pre-Calculus and Trigonometry.

Question

cos a = - 12/13, quadrant II, sin B = 1/2, quadrant II. Find cos (a-B).

Answer 1

In order to use the difference formula for cosine, we need four pieces: cos a, sin a, cos B and sin B. The problem hands us two of them, so the first job is to find the other two, and that is really what this question is testing.

The formula itself is

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

Notice the sign flips, a minus inside gives you a plus in the middle. That trips people up all the time.

Starting with a, we know cos a = -12/13, and we can find sin a from the Pythagorean identity sin^2(a) + cos^2(a) = 1. That gives sin^2(a) = 1 - 144/169 = 25/169, so sin a = 5/13 or -5/13. This is where the quadrant matters. Angle a is in quadrant II, and sine is positive there, so sin a = 5/13.

Now for B. We know sin B = 1/2, so cos^2(B) = 1 - 1/4 = 3/4 and cos B = sqrt(3)/2 or -sqrt(3)/2. B is also in quadrant II, and cosine is negative there, so cos B = -sqrt(3)/2.

Remember, our goal is cos(a - B), so now we just substitute all four values in:

cos(a - B) = (-12/13)(-sqrt(3)/2) + (5/13)(1/2)

The first piece is a negative times a negative, so it comes out positive:

cos(a - B) = 12sqrt(3)/26 + 5/26

Both already have 26 underneath, so we can just combine them.

So, our final answer is cos(a - B) = (12sqrt(3) + 5)/26, which is about 0.992.

Answer 2

In quadrant 2, cosine is negative and sine is positive

sina = √[1-cos²a] = 5/13 and cosB = -√[1-sin²B] = -√3/2

So, cos(a-B) = (cosa)(cosB) + (sina)(sinB) = (-12/13)(-√3/2) + (5/13)(1/2) = (12√3 + 5)/26