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

2. Starter: 1.Calculate the area of the shaded segment 15 cm π/4 O A B

3. θ OPPOSITEOPPOSITE H Y P O T E N U S E A D J A C E N T The three trigonometric ratios Sin θ = Opposite Hypotenuse S O H Cos θ = Adjacent Hypotenuse C A H Tan θ = Opposite Adjacent T O A 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:

4. x y O P( x, y ) r The sine, cosine and tangent of any angle 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 θ there is an associated acute angle α between the line OP and the x -axis. α θ

5. The graph of y = sin θ

6. The graph of y = cos θ

7. The graph of y = tan θ

8. 3 rd quadrant2 nd quadrant1 st quadrant4 th quadrant Tangent is positive T T Sine is positive S S All are positive A A Remember CAST We can use CAST to remember in which quadrant each of the three ratios is positive. Cosine is positive C C

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

11. 22 2 60° 2 30° 1 Sin, cos and tan of 30° Suppose we have an equilateral triangle of side length 2. We can use this triangle to write exact values for sin, cos and tan 30°: If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. Using Pythagoras’ theorem: The height of the triangle sin 30° =cos 30° =tan 30° =

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

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

14. Task 2 : Write down the value of the following leaving your answers in terms of surds where appropriate: 1.sin 120 ° 2.cos 150 ° 3.tan 225 ° 4.cos 300 ° 5.sin (-30) ° 6.cos (-120) ° 7.sin 240 ° 8.sin 420 ° 9.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 –1 k the calculator will give a value for θ between –90° and 90°. 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: There is one and only one solution in this range. Solve sin θ = 0.7 for –360° < θ < 360°. sin -1 0.7 = 44.4° (to 1 d.p.)

16. 44.4° Equations of the form sin θ = k 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. 135.6° –224.4°–315.6° Moving anticlockwise from the x -axis gives the second solution: Moving clockwise from the x -axis gives the third and fourth solutions: 180° – 44.4° = 135.6° –(180° + 44.4°) = –224.4° –(360° – 44.4°) = –315.6°

17. Examples: 1.Solve for 0 ≤ x ≤ 360°, cos x = ½ 2.Solve for 0 ≤ x ≤ 360°, sin x = - 0.685

18. Task 3: Solve for 0 ≤ x ≤ 360° 1.Sin x = 0.6 2.Cos x = 0.8 3.Tan x = 0.4 4.Sin x = -0.8 5.Cos x = -0.6 6.Tan x = -0.5

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