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

The difference formula for cos(a-b) = cosa*cosb + sina*sinb

cos a = -12/13 and sin b = 1/2 in quadrant 2 (where cosine is negative and sine is positive).

To find our answer we need to find sin a and cos b.

To determine sin a we use the fact that cos**2(a) + sin**2(a) = 1

so (-12/13)**2 + sin**2(a) = 1

144/169 + sin**2(a) = 1

sin**2(a) = 25/169

sin a = 5/13

To determine cos b we again use the fact that cos**2(b) + sin**2(b) = 1

so cos**2(b) + (1/2)**2 = 1

cos**2(b) + 1/4 = 1

cos**2(b) = 3/4

cos b = sqrt(3)/2, but since we are in 2nd quadrant cos b = -sqrt (3)/2

Now we can plug in all the values to

cos(a-b) = cosa*cosb + sina*sinb

= -12/13 * -sqrt(3)/2 + 5/13 * 1/2

= -12 * sqrt(3) / 26 + 5/26

= (-12* sqrt(3) + 5) / 26