Problem 1

1. Problem 1 • Find the exact value of cos 15° by using 15° = 60° − 45° • Once again, we use the 30°-60° and 45°-45° triangles to find the exact value. cos 15° =cos (60° − 45°) = cos 60°cos 45° + sin 60°sin 45°=

2. Problem 2 If sin α = 4/5 (in Quadrant I) and cos β = -12/13 (in Quadrant II) evaluate cos(β − α). [This is not the same as Example 2 above. This time we need to find the cosine of the difference.]

3. We use the 30-60 and 45-45 triangles again. In this case, for the cosine of the difference of two angles, we have: cos(α - β)=cos α cos β +sin α sin β= Once again, this is the exact answer.

4. Example 3 Reduce the following to a single term. Do not expand. cos(x + y)cosy + sin(x + y)sin y

5. We recognize this expression as the right hand side of: cos(α − β) = cos α cos β + cos α cos β, with α = x + y and β = y. We can thus write this in terms of cos(α − β) as follows: cos(x + y)cosy + sin(x + y)sin y = cos[(x + y) − (y)] = cosx We have reduced the expression to a single term. Note: this is the reverse of how we often do things.

6. Problem 4 Prove:

7. We recall the 30-60 triangle: Use cos(α + β) = cos α cos β − sin α sin β , with α=30° and β =x

8. Therefore….. cos(α + β) = cos α cos β − sin α sin β = =cos30°cosx-sin30°sinx. Substituting the values for 30° for sin and cos: Q.E.D.