Trigonometry Basics

1. Trigonometry Basics Module 9 Lecture 1

2. Module 9 Trigonometry • Lecture 1 • Radian measure • Trig equations • Trig identities part 1

3. Start with a circle Move a copy of the the radius and bend to lie on circumference 1 radius What is a radian?

4. 1 radius 1 radius 1 radius The angle is one radian What is a radian? Add the other radius 1 rad

5. Compare this with the equilateral triangle 1 1 1 1 1 rad 60° 1 1 Approximately how big is a radian? A radian can be though of as being constructed by taking an equilateral triangle and bending one side to form an arc so: A radian is a little bit less than 60°

6. r r 1 rad r Arc Length and Radians An arc length of 1 radius corresponds to an angle of 1 radian.

7. r r 1 rad r Arc Length and Radians An angle of 1 radian subtends an arc length of 1 radius.

8. r  r  rad r Arc Length and Radians An angle of  radians subtends an arc length of r radii

9. s r  r Hence s = r  where  is in radians Arc Length and Radians

10. Radians  Degrees r Rotate this radius through 180

11. p Semicircle has circumfere nce r p ie radii Hence this angle = p rad Thus  rad = 180 r

12. This gives the conversions: Remember by: Into radians, the radian measure goes on top Into degrees, the degree measure goes on top

13. Examples Convert 38º to radians Convert 1.34 radians to degrees Find the arc length subtended by an angle of 2.3 rads in a circle of radius 3 cm.

14. TRIGONOMETRIC EQUATIONS

15. Trig equations have the basic form The solution sets have the form, for integer n Since the trig functions are periodic, the solutions are sets

16. TRIGONOMETRIC EQUATIONS SIN

17.  -2 -  2 3 4 5 Graphically this involves finding the x-values of the intersections of y = sin x and y = 

18. -2 + arcsin  arcsin  2 + arcsin  4 + arcsin  Consider the following subset of intersections  -2 -  2 3 4 5

19.  -2 -  2 3 4 5 -2 + arcsin  arcsin  2 + arcsin  4 + arcsin  - - arcsin   - arcsin  3 - arcsin  5 - arcsin   -2 -  2 3 4 5

20. -2 + arcsin  arcsin  2 + arcsin  4 + arcsin  - - arcsin   - arcsin  3 - arcsin  5 - arcsin  We have now found all the solutions  -2 -  2 3 4 5 Pattern Each solution is of form x = n  arcsin The last term is positive for even n and negative for odd We can replace the  with (-1)n, for integer n

21.  -2 -  2 3 4 5

22. TRIGONOMETRIC EQUATIONS COS

23.  5 -2 -  2 3 4 Graphically this involves finding the x-values of the intersections of y = cos x and y = 

24.  -2 -  2 3 4 5  2 + arccos  -arccos  2 - arccos  arccos This repeats each 2 Clearly, repeating this, we can now find all the solutions

25. TRIGONOMETRIC EQUATIONS TAN

26.  - 2 3 arctan  + arctan    Tan is particularly simple as it repeats each 

27. Example Putting n=0 gives n=1 gives n=2 gives n=-1 gives etc

28. Example Note: The cos has to be solved for first.Only then can you solve for x Putting n=0 gives n=1 gives n=-1 gives etc

29. Example Note: The tan has to be solved for first.Only then can you solve for x Putting n=0 gives n=1 gives n=2 gives n=-1 gives etc

30. Example Using Degrees In this case the  in the formula needs replacing by 180 and the calculator needs setting to degrees

31. Example Using Degrees In this case the  in the formula needs replacing by 180 and the calculator needs setting to degrees

32. Example Using Degrees In this case the  in the formula needs replacing by 180 and the calculator needs setting to degrees

33. Example Using Degrees In this case the  in the formula needs replacing by 180 and the calculator needs setting to degrees

34. Example Using Degrees In this case the  in the formula needs replacing by 180 and the calculator needs setting to degrees As an answer is only required in the range 0 to 180, we can stop substituting values of n that give angles outside the range. In fact we only need n = 0, 1

35. TRIGONOMETRIC IDENTITIES Part 1

36. 1 sin  cos Identities using the trig forms of Pythagoras Use this form to get the rest Important in integration Less important

37. Prove Notice the LHS has a mixture of cos and sin terms where the RHS only has cos terms Strategy: Turn sin into cos using sin2 + cos2 = 1

38. Consider the LHS Turn sin into cos using sin2 + cos2 = 1

39. Prove Prove Prove Pythagoras: Additional Examples

40. Identities using the addition formulae

41. Example 1 Expand Strategy: Use the formula for cos(A-B)

42. Example 2 Prove LHS Strategy: You know the addition formulae for sin and cos, so replace tan by sin/cos

43. Prove LHS Strategy: Turn the sin and cos terms back into tan by dividing every term in both numerator and denominator by coscos

44. Prove LHS Strategy: Turn the sin/cos terms back into tan and cancel cos/cos terms.

45. Prove =RHS

46. Prove Prove: Prove: Addition Formulae: Additional Examples

47. Using the substitutions give us the other two forms cos2 of which are much more useful in practice Identities using the double angle formulae The addition formulae give

48. Also Identities using the double angle formulae

49. Example Prove Strategy: The RHS contains no double angles so expand the LHS double angles. Question: Which of the 3 forms to use for cos? Answer: As the RHS, being cot, must have sin in the denominator, use the form involving sin

50. Prove Simplify and cancel