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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.

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C2: Trigonometrical Equations

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  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

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