CosA = -1/6 and CosB = -3/5, with A and B in quadrant 3.?

So hopefully when you saw cos (A-B) you reflexively wrote out the difference of cosines forumula. Here's how I remember it:

cos (A-B)

cos (A + -B)

cos A cos -B - sin A sin -B

cos A cos B + sin A sin B

These mini-derivations will save you brain space. By memorizing sin (A+B) and cos (A+B), you can quickly get the difference formula (B ===> -B like I did), the double angle formula (B ===> A), and the half-angle formula (B ===> A, then A===> A/2.)

So, we need to find four things:

1) Cos A. Well, that's easy...it's given!

2) Cos B. Ditto

3) Sin A

4) Sin B

First, since they're in quadrant III, cosine and sine are both negative. (All Students Take Calculus - look up that mnemonic, but be sure to understand why it works!)

If you know cos B and want to find sin B, there are two ways to go:

Geometry way:

Draw a triangle with angle B. Label the sides so that cos B = 3/5. (Hint: SOH-CAH-TOA. Let adjacent = 3 and hypotenuse = 5. You can choose 6 and 10, but why make it harder?) Then find the remaining side with the pytagorean theorem, and then you can find the sine of angle A with opp/hypotenuse.

Algebra way:

sin^2 x + cos^2 x = 1, plug in cos x = 3/5, and solve for sin x with some rearrangement. You'll get "two" values for sin x...a number and its opposite (like -1/2 and +1/2). Pick the negative one since it's in Q III.

Once you get the four pieces of your puzzle together, you're done!

a million. My siblings, me, and my mothers and dads names all start up with D. Denise, Dennis, Devin, Daniel, break of day 2. My mothers area is Italian and Spanish, my dads is Russian 3. My mothers and dads are the two born on an identical month and day, yet not an identical 3 hundred and sixty 5 days. September seventh.

cos(A-B) = cosAcosB-sinAsinB

since A B are in Q III, sin A and sinB are -ve, sinA = -sqrt(1-cos^2A)

so sinA = -sqrt(35)/6, sinB = -4/5

cos(A-B) = 3/30 - 4*sqrt(35)/30 = -0.6888 (angle A < angle B)