1. 360° Trigonometric Graphs Higher Maths 1 2 3 Trigonometric Functions1 y = sin x Half of the vertical height. Amplitude The horizontal width of one wave section. Period Graphs of trigonometric equations are wave shaped with a repeating pattern. 720° amplitude period y = tan x Graphs of the tangent function: the amplitude cannot be measured.

2. Amplitude and Period Higher Maths 1 2 3 Trigonometric Functions2 amplitude y = 3 cos 5 x + 2 For graphs of the form y = a sin bx + c and y = a cos bx + c amplitude = a period = b 360° a = 3a = 3 period = 72° 5 360° c = 2c = 2 Example For graphs of the form y = a tan bx + c period = b 180° (amplitude is undefined)

3. Radians Higher Maths 1 2 3 Trigonometric Functions3 r r r 60° r r r one radian A radian is not 60° Angles are often measured in radians instead of degrees. A radian is the angle for which the length of the arc is the same as the radius. C = π D C = 2 π r The radius fits into the circumference times. 2 π2 π r r r r r r degrees = radians 360° 2 π2 π Radians are normally written as fractions of. π 2 π2 π ≈ 6.28… Inaccurate

4. Exact Values of Trigonometric Functions Higher Maths 1 2 3 Trigonometric Functions4 sin x x 0°30° 45°60° 90° not defined cos x tan x 0 1 0 π 2 π 3 π 4 π 6 0 2 1 2 3 √ 1 2 √ 1 3 √ 1 3 √ 2 3 √ 1 2 √ 2 1 1 0

5. Quadrants Higher Maths 1 2 3 Trigonometric Functions5 1 st 2 nd 4 th 3 rd 180°0° 90° 270° 37° 1 st 2 nd 4 th 3 rd 0 π 2 π π 2 3 π 3 4 It is useful to think of angles in terms of quadrants. 37° is in the 1 st quadrant is in the 3 rd quadrant π 3 4 Examples

6. The Quadrant Diagram Higher Maths 1 2 3 Trigonometric Functions6 1 st 2 nd 4 th 3 rd all positive sin positive tan positive cos positive 180°0° 90° 270° sin + cos + tan + sin + cos – tan – sin – cos + tan – sin – cos – tan + S A T C 1 st 2 nd 3 rd 4 th The Quadrant Diagram The nature of trigonometric functions can be shown using a simple diagram. 360°270°180°90° ++ ++

7. T + C + A + S + Quadrants and Exact Values Higher Maths 1 2 3 Trigonometric Functions7 Any angle can be written as an acute angle starting from either 0° or 180°. S + A + T + C + 120° 60° sin 120° sin 60° 2 3 √ = S + A + T + C + 225° 45° - cos 45° = 1 2 √ - = cos 225° = π 6 - tan = 1 3 √ - tan = π 6 7 - π 6 7 - π 6 cos negative tan negative

8. Higher Maths 1 2 3 Trigonometric Functions8 Solving Trigonometric Equations Graphically It is possible to solve trigonometric equations by sketching a graph. Example Solve 2 cos x – 3 = 0 for 0 x 2π2π √ 2 cos x = 2 3 √ 3 √ cos x = x = π 6 2 3 √ π 2π2π π 6 or x = π 6 2 π – = 11 π 6 6 Sketching y = cos x gives: y = cos x

9. A + C + Solving Trigonometric Equations using Quadrants Higher Maths 1 2 3 Trigonometric Functions9 Example Solve 2 sin x + 1 = 0 for 0° x 360° √ 1 2 √ sin x = - S + T +   sin negative solutions are in the 3 rd and 4 th quadrants 45° acute angle: sin - 1 ( 1 2 √ ) = 45° x = 180° + 45° = 225° x = 360° – 45° = 315° or S + A + T + C + Trigonometric equations can also be solved algebraically using quadrants. The ‘X-Wing’ Diagram

10. Higher Maths 1 2 3 Trigonometric Functions10 Example 2 Solve tan 4 x + 3 = 0 for 0 x √ π 2 tan 4 x = 3 √ - A + C + S + T +   tan negative solutions are in the 2 nd and 4 th quadrants tan - 1 () = √ 3 π 3 4 x = π 3 π – = 2π2π 3 x = π 6 or 4 x = π 3 2 π – = 5π5π 3 x = 5π5π 12 π π 2 3π3π 2 0 (continued) Solving Trigonometric Equations using Quadrants acute angle:

11. Problems involving Compound Angles Higher Maths 1 2 3 Trigonometric Functions11 Solve 6 sin ( 2 x + 10 ) = 3 for 0° x 360° Example sin ( 2 x + 10 ) = 2 1 A + C + S + T +   solutions are in the 1 st and 2 nd quadrants 30° 0° x 360° 0° 2 x 720° 10° 2 x + 10 730° Consider the range: 2 x + 10 = 30° or 150° or 390° or 510° 2 x = 20° or 140° or 380° or 500° x = 10° or 70° or 190° or 250° 360°+30° 360°+150° Don’t forget to include angles more than 360°

12. Higher Maths 1 2 3 Trigonometric Functions12 ( sin x ) 2 sin 2 x is often written Solve 7 sin 2 x + 3 sin x – 4 = 0 for 0° x 360° ( 7 sin x + 4 ) ( sin x – 1 ) = 0 7 sin x + 4 = 0 sin x – 1 = 0 or sin x = 4 - 7 sin x = 1 SA TC  acute angle ≈ 34.8° x ≈ 180° + 34.8° ≈ 214.8° or x ≈ 360° – 34.8° ≈ 325.2° x = 90° Example Solving Quadratic Trigonometric Equations