Right angled triangle

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 α b θ

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

3. Also, if we divide Sine by Cosine we get: This is aTrigonometricIdentity 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 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 is the hypotenuse (the longest side) c a θ b a2 + b2 = c2

8. Trigonometry

9. Pythagoras a2 + b2 = c2 Can be written as = 1 a2 b2 c2 c2c2c2 + =

10. proof sin θ can be written as a/c and cos θ can be written as b/c (cos) (a/c)2 is sin2θ and(b/c)2 is cos2θ (a/c)2 +(b/c)2 = 1 so sin2θ + cos2θ= 1 (sin2 θ) means to find the sine of θ, then square it. (sin θ2) means square θ, then find the sine

11. Rearranged versions sin2θ = 1 − cos2θ cos2θ = 1 − sin2θ

12. Rearranged versions sin2θ cos2θ 1 cos2θ cos2θ cos2θ tan2θ + 1 = sec2θ tan2θ = sec2θ − 1 Or sec2θ =1 + tan2θ + = OR

13. Rearranged versions sin2θ cos2θ1 sin2θsin2θsin2θ 1 + cot2θ = cosec2θ cot2θ +1 = cosec2θ or cosec2θ =1 + cot2θ = +

14. Circular motion to sine curve y = R.sinθ R y θ time 0o 90o 180o 270o 360o 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.sinθ. At 90o sinθ = 1 so y = R

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

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

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

18. Graph of sin2θ Has to be positive because we cannot have a minus squared number

19. Graph of cos2θ

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

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

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

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

24. Finding angles Find all the angles between 0o and 360o to satisfy the equation 8sinθ -4 = 0 (rearrange) 8sinθ = 4 Sinθ = 4/8 = 0.5 Sin-10.5 = 30o and 180o – 30o = 150o

25. Find all the angles between 0o and 360o to satisfy the equation 6cos2θ = 1.8 (rearrange) cos2θ= 1.8÷6 = 0.3 cosθ= √0.3 = ± 0.548 Cos-1 +0.548 (1st and 4th quadrant positive) = 56.8o and 360o – 56.8o = 303.2o Cos-1- 0.548 (2ndand 3rdquadrant negative) 180o – 56.8o = 123.2o and 180o + 56.8o = 236.8o

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

27. Finding angles • Solve for all angle between 0° and 360° • 2Tan2 B + Tan B = 6 • (let Tan B = x)so • 2x2+ x = 6 or 2x2+ x – 6 = 0 • then solving as a quadratic equation using formula: • x = -b +/- √(b2 - 4ac) / 2a • Where a = 2; b= 1; and c = - 6

28. Finding angles • x = -1+/- √(12 – 4x2x-6) / 4 • = -1+/- √(1+ 48) / 4 = -1+/- √(49) / 4 = -1+/- (7) / 4 +6/4 or -8/4 Tan B = 1.5 or -2 1st and 3rd quadrant 56.3o or(180 + 56.3) = 236.3o 2nd quadrant (180 - 63.43) = 116.57o • 4th quadrant (360 – 63.43) = 296.57o

29. Formulaefor sin (A + B), cos (A + B), tan (A + B) (compoundangles) • 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 • Thesewill come in handylater

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 Rsin (θ + α) as follows: • R sin (θ + α) •  ≡ R (sin θcosα + cosθsin α) •  ≡ R sin θcosα + Rcosθsin α

32. Finding α • So asin θ + bcosθ≡ R cosαsin θ + R sin αcosθ • a = R cosα • b = R sin α

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

34. Using the equation • Now we square the equation • a2 + b2 • = R2cos2 α + R2 sin2 α • = R2(cos2 α + sin2 α) • = R2 • (because cos2 A + sin2 A = 1)

35. compound angle formulae: • Hence • R =√a2+ b2 • R2 = a2 + b2 • (pythagoras)

36. The important bits tan α = b/a • R2 = a2 + b2 • (pythagoras)

37. For the minus case • a sin θ − bcosθ • = • Rsin(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/afor the sine  case.) • We find R the same as before: • R=√a2 +b2 • So the sum of a sine term and cosine term have been combined into a single cosine term: • a sin θ + bcosθ ≡ Rcos(θ − α)

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

40. Graph of 4sinθ

41. Graph of 4sinθ and 3 cosθ

42. Resultant graph of 4sinθ+ 3 cosθ

43. The radian 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 r That is 3600 = 2π radians

44. Circular motion to sine curve y = R.sinθ R y θ time Radians π 0 π/2 2π 3π/2 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.sinθ. At π/2 sinθ = 1 so y = R

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=√32+62 R =√9 +36 R =√45 R = 6.7 Maximum value is 6.7

48. example Tan α = b/a Tan α = 6/3 = 2 63.4o or 1.107 radians 63.4o 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 • 3Sin ωt - 6Cos ωt = 6.7Sin (ωt - 1.107) • maximum = 6.7 • Maximum value occurs at 2.6781 radians