Outline of Chapter 5 (for now)

1. Outline of Chapter 5 (for now) • Derive formulas sums and differences of trigonometric functions: cos(α ± β) sin (α ± β), and tan (α ± β) • To prove, we need to know: odd and even functions for cos(α ± β) complementarity of angles for sin (α ± β)

2. Sum and differences of two angles The sine of the sum and difference of two angles is as follows: sin(α + β) = sin α cosβ + cosα sin β sin(α − β) = sin α cosβ − cosα sin β The cosine of the sum and difference of two angles is as follows: cos(α + β) = cosα cosβ − sin α sin β cos(α − β) = cosα cosβ + sin α sin β

3. Proof of cos(α+β) We will prove the cosine of the sum of two angles identity first, and then show that this result can be extended to all the other identities given. PROVE: cos (α+β) = cos α cos β − sin α sin β

4. Drawing the unit circle We draw an angle α from the centre with terminal point Q at (cos α, sin α), as shown. [Q is (cos α, sin α) because the hypotenuse is 1 unit.]

5. We extend this idea by drawing: • The angle β with terminal points at Q (cosα, sin α) and R (cos (α + β), sin (α + β)) • The angle -β with terminal point at S (cos (-β), sin (-β)) • The lines PR and QS, which are equivalent in length.

6. Now, using the Pythagorean theorem, we have: PR2 = (cos (α + β) − 1)2 + sin2(α + β) = cos2(α + β) − 2cos (α + β) + 1 + sin2(α + β) = 2 − 2cos (α + β) [since sin2(α + β) + cos2(α + β) = 1]

7. Now using the same approach on QS: QS2 = (cosα − cos (-β))2 + (sin α − sin (-β))2 = cos2α − 2cosα cos(-β) + cos2(-β) + sin 2α − 2sin α sin(-β) + sin2(-β) = 2 − 2cosα cos(-β) − 2sin α sin(-β) [since sin2α + cos2α = 1 and sin2(-β) + cos2(-β) = 1]

8. Here, we need to review odd and even functions Even function (symmetric with respect to y-axis): Odd function (symmetric with respect to origin):

9. Applying the odd and even function concept: 2 − 2cosα cos(-β) − 2sin α sin(-β) (where we left off) Substitute cos(-β) = cosβ (cosine is an even function) and sin(-β) = -sinβ (sin is an odd function) = 2 − 2cosα cosβ + 2sin α sin β

10. Since PR = QS, we can equate the 2 distances we just found: 2 − 2cos (α + β) = 2 − 2cosα cosβ + 2sin α sin β Subtracting 2 from both sides and dividing throughout by -2, we obtain: cos (α + β) = cos α cos β − sin α sin β

11. But what about cos (α - β)? • If we replace β with (-β), this identity— cos (α + β) = cos α cos β − sin α sin β—becomes: cos (α − β) = cosα cosβ + sin α sin β [since cos(-β) = cosβ and sin(-β) = -sinβ]

12. On to sin (α ± β) But first…… a brief diversion to the complementarily of the trig functions

13. Complementarity Consider a standard right triangle: How are cos A and sin B related? cos A = b/c. So does sin B……

14. What about cos B and sin B? cos B = a/c. So does sin A…… Conclusion: in a right triangle, cos A = sin (90° –A) = sin (π/2-A).

15. Verbal phrasings/definition The cosine of any acute angle is equal to the sine of the compliment, and vice versa Alternative phrasings: cos A = sin (90° –A) = sin (π/2-A) sin A = cos (90° –A) = cos (π/2-A)

16. Homework • Using this triangle, determine the complimentary relationships for tan A, cot A, sec A, and csc A

17. Application to finding sin (α ± β) Goal: prove sin (α + β) = sin α cos β + cos α sin β We recall that sin θ = cos (π/2− θ) (1) Set θ= α + β and substituteintoequation (1): sin (α + β)= cos [π/2 − (α + β)] = cos [π/2 − α − β)]

18. Regroup within the parentheses: cos [π/2 − α − β)] = cos [(π/2 − α) − β] Now apply earlier definition of cos (α−β): cos [(π/2 − α) − β] = cos (π/2 − α) cos (β) + sin (π/2 − α) sin (β) = sin α cos β + cos α sin β [Since cos (π/2 − α) = sin α; and sin (π/2 − α) = cosα] Therefore, sin(α + β) = sin α cos β + cos α sin β

19. sin (α − β) is derived like cos (α − β) Replace β with − β in sin α cos β + cos α sin β : sin(α +(− β)) = sin α cos (-β) + cos α sin(-β) Nowapplyodd and evenproperties of sin and costoget: sin(α − β) = sin α cos (-β) + cos α sin(-β) = sin α cos (β) - cos α sin(β). Therefore, sin(α − β) = sin α cos β - cos α sinβ

20. Tangent is a bit more complex….

21. The trig identities strike back….. Or, at least, so does the quotient property of tanα: tanα = sinα/cosα Substitute this identity for tan(α+β):

22. And now, a trick…… Divide top and bottom by cosαcosβ:

23. Rewrite and start cancelling: which reduces to:

24. By now you recognize the familiar form of and the complicated fraction reduces to: Q.E.D.

25. tan(α − β) is the same old story…. tanθ is an odd function (in oh, so many ways), so therefore tan(-θ)=-tan(θ). We substitute − β for β in the tan (α + β) equation:

26. Example 1 Find the exact value of cos 75° by using 75° = 30° + 45°. Begin by recalling the 30-60-90 and 45-45-90 triangles:

27. We use the exact sine and cosine ratios from the triangles to answer the question as follows: • This is the exact value for cos 75°.

28. Example 2 If sin α = 4/5 (in Quadrant I) and cosβ = -12/13 (in Quadrant II) evaluate sin(α − β). We use sin(α − β) = sin α cos β − cos α sin β We first need to find cos α and sin β.

29. If sin α = 4/5, then we can draw a triangle and find the value of the unknown side using the Pythagorean Theorem (in this case, 3): Same for cos β = 12/13:

30. Now for the unknown ratios in the question: cosα = 3/5 (positive because in quadrant I) sin β = 5/13 (positive because in quadrant II) sin(α − β) = sin α cosβ − cosα sin β= This is the exact value for sin(α − β).