1 / 21

C2: Trigonometrical Equations

C2: Trigonometrical Equations. Learning Objective: to be able to solve simple trigonometrical equations in a given range. A. π /4. B. O. 15 cm. Starter:. Calculate the area of the shaded segment. The three trigonometric ratios. H. Y. P. O. O P P O S I T E. T. E. N. U. S.

Download Presentation

C2: Trigonometrical Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. C2: Trigonometrical Equations Learning Objective: to be able to solve simple trigonometrical equations in a given range

  2. A π/4 B O 15 cm Starter: • Calculate the area of the shaded segment

  3. The three trigonometric ratios H Y P O O P P O S I T E T E N U S Opposite Adjacent Opposite E Cos θ= Sin θ= Tan θ= Hypotenuse Hypotenuse Adjacent θ A D J A C E N T Remember: S O H C A H T O A The three trigonometric ratios, sine, cosine and tangent, can be defined using the ratios of the sides of a right-angled triangle as follows: S O H C A H T O A

  4. The sine, cosine and tangent of any angle y P(x, y) r θ α O x These definitions are limited because a right-angled triangle cannot contain any angles greater than 90°. To extend the three trigonometric ratios to include angles greater than 90° and less than 0° we consider the rotation of a straight line OP of fixed length r about the origin O of a coordinate grid. Angles are then measured anticlockwise from the positive x-axis. For any angle θthere is an associated acute angle α between the line OP and the x-axis.

  5. The graph of y = sin θ

  6. The graph of y = cos θ

  7. The graph of y = tan θ

  8. 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 is positive.

  9. Task 1: Write each of the following as trigonometric ratios of positive acute angles: • sin 260° • cos 140° • tan 185° • tan 355° • cos 137° • sin 414° • sin (-194)° • cos (-336)° • tan 396° • tan 148°

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

  11. Sin, cos and tan of 30° 60° 30° 2 2 2 The height of the triangle 60° 60° 60° 1 2 sin 30° = cos 30° = tan 30° = 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°. Using Pythagoras’ theorem: We can use this triangle to write exact values for sin, cos and tan 30°:

  12. Sin, cos and tan of 60° 60° 30° 2 2 2 The height of the triangle 60° 60° 60° 1 2 sin 60° = tan 60° = cos 60° = 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°. Using Pythagoras’ theorem: We can also use this triangle to write exact values for sin, cos and tan 60°:

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

  14. Task 2 : Write down the value of the following leaving your answers in terms of surds where appropriate: • sin 120° • cos 150° • tan 225° • cos 300° • sin (-30)° • cos (-120)° • sin 240° • sin 420° • cos 315°

  15. Equations of the form sin θ= k Equations of the form sin θ= k, where –1 ≤ k ≤ 1, have an infinite number of solutions. If we use a calculator to find sin–1k the calculator will give a value for θ between –90° and 90°. There is one and only one solution in this range. This is called the principal solution of sin θ= k. Other solutions in a given range can be found by considering the unit circle. For example: Solve sin θ = 0.7 for –360° < θ < 360°. sin-1 0.7 = 44.4° (to 1 d.p.)

  16. Equations of the form sin θ= k 44.4° We solve sin θ = 0.7 for –360° < θ < 360° by considering angles in the first and second quadrants of a unit circle where the sine ratio is positive. Start by sketching the principal solution 44.4°in the first quadrant. Next, sketch the associated acute angle in the second quadrant. –224.4° –315.6° Moving anticlockwise from thex-axis gives the second solution: 135.6° 180° – 44.4° = 135.6° Moving clockwise from thex-axis gives the third and fourth solutions: –(180° + 44.4°) = –224.4° –(360° – 44.4°) = –315.6°

  17. Examples: • Solve for 0 ≤ x ≤ 360°, cos x = ½ • Solve for 0 ≤ x ≤ 360°, sin x = - 0.685

  18. Task 3: Solve for 0 ≤ x ≤ 360° • Sin x = 0.6 • Cos x = 0.8 • Tan x = 0.4 • Sin x = -0.8 • Cos x = -0.6 • Tan x = -0.5

  19. Task 4: Solve for 0 ≤ x ≤ 2π • sin x = 1/2 • cos x = 1/ √2 • tan x = - √3 • sin x = √3 / 2 • cos x = 1/2 • cos x = - 1 / √2

  20. Examples: Solve for -180° ≤ x ≤ 180°, tan 2x = 1.424 Solve for 0 ≤ x ≤ 360°, sin (x + 30°) = 0.781

  21. Task 5: Solve for 0 ≤ x ≤ 360° • sin 3x = 0.7 • sin (x / 3) = 2/3 • tan 4x = 1/3 • cos 2x = 0.63 • cos (x + 72°) = 0.515 Solve for 0 ≤ x ≤ 2π • tan 2x = 1 • sin (x / 3) = ½ • sin (x + π/6) = √3 / 2

More Related