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KS4 Mathematics

KS4 Mathematics. S4 Further trigonometry. S4 Further trigonometry. Contents. A. S4.2 Sin, cos and tan of 30°, 45° and 60°. A. S4.3 Graphs of trigonometric functions. A. S4.1 Sin, cos and tan of any angle. S3.4 Area of a triangle using ½ ab sin C. A. S3.5 The sine rule. A.

harlan-briggs
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KS4 Mathematics

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  1. KS4 Mathematics S4 Further trigonometry

  2. S4 Further trigonometry Contents • A S4.2 Sin, cos and tan of 30°, 45° and 60° • A S4.3 Graphs of trigonometric functions • A S4.1 Sin, cos and tan of any angle S3.4 Area of a triangle using ½ab sin C • A S3.5 The sine rule • A S4.6 The cosine rule • A

  3. The opposite and adjacent sides h θ a) Write an expression for the length of the opposite side in terms of h and θ. b) Write an expression for the length of the adjacent side in terms of h and θ. Suppose we have a right-angled triangle with hypotenuse h and acute angle θ.

  4. The opposite and adjacent sides h θ opp adj a) sin θ = b) cos θ = hyp hyp Suppose we have a right-angled triangle with hypotenuse h and acute angle θ. opp = hyp × sin θ adj = hyp × cos θ opp = h sin θ adj = h cos θ

  5. The opposite and adjacent sides h sin θ tan θ = h cos θ sin θ tan θ = cos θ So, for any right-angled triangle with hypotenuse h and acute angle θ. We can label the opposite and adjacent sides as follows: h h sin θ θ h cos θ opposite We can write, adjacent

  6. The sine of any angle

  7. Sine of angles in the second quadrant We have seen that the sine of angles in the first and second quadrants are positive. The sine of angles in the third and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. sin θ = sin (180° – θ) For example, sin 130° = sin (180° – 130°) = sin 50° = 0.766 (to 3 sig. figs)

  8. Sine of angles in the third quadrant In the third quadrant, 180° < θ < 270°. sin θ = –sin (θ– 180°) For example, sin 220° = – sin (220° – 180°) = – sin 40° = – 0.643 (to 3 sig. figs) Verify, using a scientific calculator, that sin 220° = –sin 40°

  9. Sine of angles in the fourth quadrant In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° sin θ = –sin(360° – θ) or sin –θ = –sinθ For example, sin 300° = –sin (360° – 300°) = –sin 60° = –0.866 (to 3 sig. figs) sin –35° = –sin 35° = –0.574 (to 3 sig. figs)

  10. The cosine of any angle

  11. Cosine of angles in the second quadrant We have seen that the cosines of angles in the first and fourth quadrants are positive. The cosines of angles in the second and third quadrants are negative. In the second quadrant, 90° < θ < 180°. cos θ = –cos (180° – θ) For example, cos 100° = –cos (180° – 100°) = –cos 80° = –0.174 (to 3 sig. figs)

  12. Cosine of angles in the third quadrant In the third quadrant, 180° < θ < 270°. cos θ = –cos (θ– 180°) For example, cos 250° = –cos (250° – 180°) = –cos 70° = –0.342 (to 3 sig. figs.) Verify, using a scientific calculator, that cos 250° = –cos 70°

  13. Sine of angles in the fourth quadrant In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° cos θ = cos(360° – θ) or cos –θ = cosθ For example, cos 317° = cos (360° – 317°) = cos 43° = 0.731 (to 3 sig. figs.) cos –28° = cos 28° = 0.883 (to 3 sig. figs.)

  14. The tangent of any angle

  15. The tangent of any angle

  16. Tangent of angles in the second quadrant We have seen that the tangent of angles in the first and third quadrants are positive. The tangent of angles in the second and fourth quadrants are negative. In the second quadrant, 90° < θ < 180°. tan θ = –tan (180° – θ) For example, tan 116° = –tan (180° – 116°) = –tan 64° = –2.05 (to 3 sig. figs)

  17. Tangent of angles in the third quadrant In the third quadrant, 180° < θ < 270°. tan θ = tan (θ– 180°) For example, tan 236° = tan (236° – 180°) = tan 56° = 1.48 (to 3 sig. figs) Verify, using a scientific calculator, that tan 236° = tan 56°

  18. Tangent of angles in the fourth quadrant In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° tan θ = –tan(360° – θ) or tan –θ = –tanθ For example, tan 278° = –tan (360° – 278°) = –tan 82° = –7.12 (to 3 sig. figs) tan –16° = –tan 16° = –0.287 (to 3 sig. figs)

  19. Sin, cos and tan of angles between 0° and 360° In the second quadrant: In the third quadrant: In the fourth quadrant: The sin, cos and tan of angles in the first quadrant are positive. sin θ= sin (180° – θ) cos θ= –cos (180° – θ) tan θ= –tan (180° – θ) sin θ= –sin (θ– 180°) cos θ= –cos (θ– 180°) tan θ= tan (θ– 180°) sin θ= –sin (360° – θ) cos θ= cos (360° – θ) tan θ= –tan(180° – θ)

  20. Remember CAST 2nd quadrant 1st quadrant S Sine is positive A All are positive 3rd quadrant 4th quadrant T Tangent is positive C Cosine is positive We can use CAST to remember in which quadrant each of the three ratios are positive.

  21. Positive or negative?

  22. Find the equivalent ratio

  23. Solving equations in θ

  24. S4 Further trigonometry Contents S4.1 Sin, cos and tan of any angle • A • A S4.3 Graphs of trigonometric functions • A S4.2 Sin, cos and tan of 30°, 45° and 60° S3.4 Area of a triangle using ½ab sin C • A S3.5 The sine rule • A S4.6 The cosine rule • A

  25. Sin, cos and tan of 45° The hypotenuse =  1² + 1² 1 1 sin 45° = cos 45° = tan 45° = 2 2 A right-angled isosceles triangle has two acute angles of 45°. Suppose the equal sides are of 1 unit length. 45° 1 2 Using Pythagoras’ theorem, 45° = 2 1 We can use this triangle to write exact values for sin, cos and tan 45°: 1

  26. Sin, cos and tan of 30° 60° 30° 2 2 2 The height of the triangle =  2² – 1² 60° 60° 60° 1 2 3 1 1 sin 30° = cos 30° = tan 30° = 3 2 2 Suppose we have an equilateral triangle of side length 2. If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, = 3 We can use this triangle to write exact values for sin, cos and tan 30°:

  27. Sin, cos and tan of 60° 30° 2 The height of the triangle =  2² – 1² 60° 1 3 1 sin 60° = cos 60° = tan 60° = 2 2 Suppose we have an equilateral triangle of side length 2. If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. 3 Using Pythagoras’ theorem, = 3 We can also use this triangle to write exact values for sin, cos and tan 60°: 3

  28. Sin, cos and tan of 30°, 45° and 60° 30° 45° 60° 1 sin 1 1 1 2 2 2 2 1 1 cos 3 3 2 2 2 3 tan The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 3 1 Use this table to write the exact value of sin 150°: sin 150° =

  29. Sin, cos and tan of 30°, 45° and 60° 30° 45° 60° 1 sin 1 1 2 2 2 1 1 cos 3 3 2 2 2 3 tan –1 2 The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 3 1 Use this table to write the exact value of cos 135°: cos 135° =

  30. Sin, cos and tan of 30°, 45° and 60° 30° 45° 60° 1 sin 1 1 2 2 2 1 1 cos 3 3 2 2 2 3 tan The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: 3 1 Use this table to write the exact value of tan 120° tan 120° = –3

  31. Sin, cos and tan of 30°, 45° and 60° 1 1 2 2 1 3 2 2 –1 2 Write the following ratios exactly: 1) cos 300° = 2) tan 315° = –1 3) tan 240° = 3 4) sin –330° = 5) cos –30° = 6) tan –135° = 1 7) sin 210° = 8) cos 315° =

  32. S4 Further trigonometry Contents S4.1 Sin, cos and tan of any angle • A S4.2 Sin, cos and tan of 30°, 45° and 60° • A • A S4.3 Graphs of trigonometric functions S3.4 Area of a triangle using ½ab sin C • A S3.5 The sine rule • A S4.6 The cosine rule • A

  33. The graph of sin θ

  34. The graph of cos θ

  35. The graph of tan θ

  36. Transforming trigonometric graphs

  37. S4 Further trigonometry Contents S4.1 Sin, cos and tan of any angle • A S4.2 Sin, cos and tan of 30°, 45° and 60° • A S4.3 Graphs of trigonometric functions • A S3.4 Area of a triangle using ½ab sin C • A S3.5 The sine rule • A S4.6 The cosine rule • A

  38. The area of a triangle 1 Area of a triangle = bh 2 Remember, h b

  39. The area of a triangle A 4 cm 47° C B = sin 47° 7 cm h 4 Suppose that instead of the height of a triangle, we are given the base, one of the sides and the included angle. For example, What is the area of triangle ABC? Let’s call the height of the triangle h. We can find h using the sine ratio. h h = 4 sin 47° Area of triangle ABC = ½ × base × height = ½ × 7 × 4 sin 47° = 10.2 cm2 (to 1 d.p.)

  40. The area of a triangle using ½ ab sin C 1 Area of triangle ABC = ab sin C 2 The area of a triangle is equal to half the product of two of the sides and the sine of the included angle. A c b B C a

  41. The area of a triangle using ½ ab sin C

  42. S4 Further trigonometry Contents S4.1 Sin, cos and tan of any angle • A S4.2 Sin, cos and tan of 30°, 45° and 60° • A S4.3 Graphs of trigonometric functions • A S3.5 The sine rule S3.4 Area of a triangle using ½ab sin C • A • A S4.6 The cosine rule • A

  43. The sine rule h h a b Consider any triangle ABC, If we drop a perpendicular line, h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC. C b a h a is the side opposite A and b is the side opposite B. A B D sin B = sin A = h = a sin B h = b sin A b sin A = a sin B So,

  44. The sine rule a c b b = = sin A sin C sin B sin B b sin A = a sin B Dividing both sides of the equation by sin A and then by sin B we have: If we had dropped a perpendicular from A to BC we would have found that: b sin C = c sin B Rearranging:

  45. The sine rule C b a A B c b c sin A sin B sin C a = = = = sin B sin C a b c sin A For any triangle ABC, or

  46. Using the sine rule to find side lengths B 39° 7 a = a sin 39° sin 118° 118° 7 sin 118° a = C A 7 cm sin 39° If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine rule to find the length of the side opposite the other angle. For example, Find the length of side a Using the sine rule, a = 9.82 (to 2 d.p.)

  47. Using the sine rule to find side lengths

  48. Using the sine rule to find angles C sin 46° sin B = 8 cm 8 6 6 cm 8 sin 46° sin B = 6 8 sin 46° 46° A B sin–1 B = 6 If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine rule to find the angle opposite the other given side. For example, Find the angle at B Using the sine rule, B= 73.56° (to 2 d.p.)

  49. Finding the second possible value C 8 cm 6 cm 6 cm 46° 46° A B B Suppose that in the last examplewe had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°. There is a second possible value for the angle at B. Instead of this triangle … … we could have this triangle. Remember, sin θ= sin (180° – θ) So for every acute solution, there is a corresponding obtuse solution. B= 73.56° (to 2 d.p.) or B= 180° – 73.56° = 106.44° (to 2 d.p.)

  50. Using the sine rule to find angles

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