1. Right angled triangle C is the hypotenuse (Always the longest side) For angle θ (a) is the opposite and (b )is the adjacent For angle α (b) is the opposite and (a) is the adjacent c a b θ α

2. Trigonometry Functions Sine = opposite hypotenuse Cosine = adjacent hypotenuse Tangent = opposite adjacent

3. This is aTrigonometric Identity Also, if we divide Sine by Cosine we get: Also, if we divide Sine by Cosine we get: Sin θ = Opposite/hypotenuse Cos θ = Adjacent/hypotenuse The hypotenuses cancel each other out] So we get Opposite/adjacent which is Tan (tangent) so Tan θ = sin θ cos θ

4. More functions Cosecant Function: csc(θ) = Hypotenuse / Opposite Secant Function: sec(θ) = Hypotenuse / Adjacent Cotangent Function: cot(θ) = Adjacent / Opposite We can also divide "the other way around" (such as hypotenuse/opposite instead of Opposite/hypotenuse): which will give us three more functions

5. So using the inverse of these we get: sin(θ) = 1/cosec(θ) cos(θ) = 1/sec(θ) tan(θ) = 1/cot(θ)

6. More functions Also the other way around: cosec(θ) = 1/sin(θ) sec(θ) = 1/cos(θ) cot(θ) = 1/tan(θ) And we also have: (cot(θ) = cos(θ)/sin(θ) (from tan = sin/cos)

7. Pythagoras c a b a 2 + b 2 = c 2 C is the hypotenuse (the longest side) θ

8. TRIGONOMETRY

9. Pythagoras a 2 + b 2 = c 2 Can be written as a 2 b 2 c 2 c 2 c 2 c 2 + = = 1

10. proof sin θ can be written as a/c and cos θ can be written as b/c (cos) (a/c) 2 is sin 2 θ and(b/c) 2 is cos 2 θ (a/c) 2 +(b/c) 2 = 1 so sin 2 θ + cos 2 θ = 1 (sin 2 θ) means to find the sine of θ, then square it. (sin θ 2 ) means square θ, then find the sine

11. Rearranged versions sin 2 θ = 1 − cos 2 θ cos 2 θ = 1 − sin 2 θ

12. Rearranged versions sin 2 θ cos 2 θ 1 cos 2 θ cos 2 θ cos 2 θ tan 2 θ + 1 = sec 2 θ tan 2 θ = sec 2 θ − 1 Or sec 2 θ = 1 + tan 2 θ + = OR

13. Rearranged versions sin 2 θ cos 2 θ 1 sin 2 θ sin 2 θ sin 2 θ 1 + cot 2 θ = cosec 2 θ cot 2 θ + 1 = cosec 2 θ or cosec 2 θ = 1 + cot 2 θ + =

14. Circular motion to sine curve 0o0o 90 o 180 o 270 o 360 o time y = R.sinθ In the triangle formed by the first part of the motion. The vertical line (y) is the opposite of the angle formed (θ) and the hypotenuse is the radius of the circle (R) Sinθ = y/R so y = R.sin θ. At 90 o sinθ = 1 so y = R θ y R

15. More sine curves y = Rsinθ y =2Rsinθ y =0.5Rsinθ

16. y = sin2θ 90 o 180 o 270 o 360 o For y = sin2θ two waves fit in 360 o For y = sin3θ three waves fit in 360 o and so on For y =sin 0.5θ one wave would stretch over 720 o

17. Cosine curves (cosine 90 o = 0) (cosine 0 o =1)

18. Graph of sin 2 θ Has to be positive because we cannot have a minus squared number

19. Graph of cos 2 θ

20. C AS T A = All positive S = Only sine positive T = Only tangent positive C = only cosine positive 0o0o 90 o 180 o 270 o 360 o C.A.S.T

21. Finding sin, cos and tan of angles. Sin 245 o = sin(245 o – 180 o ) = sin 65 o = 0.906 Sin 245 o = - 0.906 (third quadrant = negative) Sin 118 o = sin (180 o – 118 o ) = sin 62 o = 0.883 Sin 118 o = + 0.883 (second quadrant positive) Cos 162 o = cos (180 o – 162 o ) = cos 18 o = 0.951 Cos 162 o = - 0.851 (second quadrant negative) Cos 285 o = cos(360 o – 285 o ) = cos 75 o = 0.259 Cos 285 o = + 0.259 (fourth quadrant positive) Tan 196 o = tan(196 o – 180 o ) = tan 16 o = 0.287 Tan 196 o = + 0.287 (third quadrant positive) Tan 282 o = tan(360 o – 282 o ) = tan 78 o = 4.705 Tan 282 o = - 4.705 (fourth quadrant negative)

22. Finding angles First quadrant θ 2 nd quadrant 180 – θ 3 rd quadrant 180 + θ 4 th quadrant 360 – θ

23. Finding angles 90 o 180 o 270 o 360 o 0o0o 30 o 180 o -30 o = 150 o 180 o + 30 o =210 o 360 o - 30 o =330 o 150 o 210 o 330 o C A S T

24. Finding angles Find all the angles between 0 o and 360 o to satisfy the equation 8sinθ -4 = 0 (rearrange) 8sinθ = 4 Sinθ = 4/8 = 0.5 Sin -1 0.5 = 30 o and 180 o – 30 o = 150 o

25. Find all the angles between 0 o and 360 o to satisfy the equation 6cos 2 θ = 1.8 (rearrange) cos 2 θ = 1.8÷6 = 0.3 cosθ = √0.3 = ± 0.548 Cos -1 +0.548 (1 st and 4 th quadrant positive) = 56.8 o and 360 o – 56.8 o = 303.2 o Cos -1 - 0.548 (2 nd and 3 rd quadrant negative ) 180 o – 56.8 o = 123.2 o and 180 o + 56.8 o = 236.8 o

26. Finding angles 90 o 180 o 270 o 360 o 0o0o 56.8 o 123.2 o 56.8 o 236.8 o 303.2 o Red 1 st and 4 th quadrant (positive cos) Blue 2 nd and 3 rd quadrant (negative cos)

27. Finding angles Solve for all angle between 0° and 360° 2Tan 2 B + Tan B = 6 (let Tan B = x)so 2x 2 + x = 6 or 2x 2 + x – 6 = 0 then solving as a quadratic equation using formula: x = -b +/- √(b 2 - 4ac) / 2a Where a = 2; b= 1; and c = - 6

28. Finding angles x = -1+/- √(1 2 – 4x2x-6) / 4 = -1+/- √(1 + 48) / 4 = -1+/- √(49) / 4 = -1+/- (7) / 4 +6/4 or -8/4 Tan B = 1.5 or -2 1 st and 3 rd quadrant 56.3 o or(180 + 56.3) = 236.3 o 2 nd quadrant (180 - 63.43) = 116.57 o 4 th quadrant (360 – 63.43) = 296.57 o

29. Formulae for sin (A + B), cos (A + B), tan (A + B) (compound angles) sin (A + B) = sin A cos B + cos A sin B sin (A - B) = sin A cos B - cos A sin B cos (A + B) = cos A cos B - sin A sin B cos (A - B) = cos A cos B + sin A sin B These will come in handy later

30. a sin θ ± b cos θ can be expressed in the form R sin(θ ± α), R is the maximum value of the sine wave sin(θ ± α) must = 1 or -1 (α is the reference angle for finding θ)

31. Finding α Using sin(A + B) = sin A cos B + cos A sin B, (from before) we can expand R sin (θ + α) as follows: R sin (θ + α) ≡ R (sin θ cos α + cos θ sin α) ≡ R sin θ cos α + R cos θ sin α

32. Finding α So a sin θ + b cos θ ≡ R cos α sin θ + R sin α cos θ a = R cos α b = R sin α

33. Finding α b ÷ a = R sin α ÷ R cos α = tan α tan α = b/a

34. Using the equation Now we square the equation a 2 + b 2 = R 2 cos 2 α + R 2 sin 2 α = R 2 (cos 2 α + sin 2 α) = R 2 (because cos 2 A + sin 2 A = 1)

35. compound angle formulae: Hence R = √a 2 + b 2 R 2 = a 2 + b 2 (pythagoras)

36. The important bits tan α = b/a R 2 = a 2 + b 2 (pythagoras)

37. For the minus case a sin θ − b cos θ = R sin(θ − α) tan α = b/a

38. Cosine version a sin θ + b cos θ ≡ R cos (θ − α) Therefore: tanα= a/b (Note the fraction is a/b for the cosine case, whereas it is b/a for the sine case.) We find R the same as before: R=√a 2 +b 2 So the sum of a sine term and cosine term have been combined into a single cosine term: a sin θ + b cos θ ≡ R cos(θ − α)

39. Minus cosine version If we have a sin θ − b cos θ and we need to express it in terms of a single cosine function, the formula we need to use is: a sin θ − b cos θ ≡ −R cos (θ + α)

40. Graph of 4sinθ

41. Graph of 4sinθ and 3 cosθ

42. Resultant graph of 4sinθ + 3 cosθ

43. The radian r Length of arc (s) The radian is the length of the arc divided by the radius of the circle, A full circle is 2π radians That is 360 0 = 2π radians

44. Circular motion to sine curve π/2 π Radians 3π/2 2π2π time y = R.sinθ In the triangle formed by the first part of the motion. The vertical line (y) is the opposite of the angle formed (θ) and the hypotenuse is the radius of the circle (R) Sinθ = y/R so y = R.sin θ. At π/2 sinθ = 1 so y = R θ y R 0

45. Angular velocity (ω) Angular velocity is the rate of change of an angle in circular motion and has the symbol ω ω = radians ÷ time (secs) Angles can be expressed by ωt

46. example For the equation 3.Sin ωt - 6.Cos ωt: i) Express in R.Sin (ωt - α) form ii) State the maximum value iii) Find value at which maximum occurs

47. example R=√3 2 +6 2 R =√9 +36 R =√45 R = 6.7 Maximum value is 6.7

48. example Tan α = b/a Tan α = 6/3 = 2 63.4 o or 1.107 radians 63.4 o x π ÷ 180

49. example Maximum value occurs when Sin (ωt - 1.1071) = 1, or (ωt - 1.1071) = π/2 radians Since π/2 radians = 1.57, then (ωt - 1.107) = 1.57 rad. Therefore ωt = 1.57 + 1.107 = 2.6781 radians Maximum value occurs at 2.678 radians

50. example a) 3Sin ω t - 6Cos ω t = 6.7Sin ( ω t - 1.107) b) maximum = 6.7 c) Maximum value occurs at 2.6781 radians