Successfully reported this slideshow.
Upcoming SlideShare
×

# Higher Maths 1.2.3 - Trigonometric Functions

15,581 views

Published on

Published in: Technology, Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Wow!

Are you sure you want to  Yes  No
• Good Job.Keep up the good work

Are you sure you want to  Yes  No

### Higher Maths 1.2.3 - Trigonometric Functions

1. 1. 360° Trigonometric Graphs NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART x y 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 x y Graphs of the tangent function: the amplitude cannot be measured.
2. 2. Amplitude and Period NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART x y 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 = 3 period = 72° 5 360° c = 2 Example For graphs of the form y = a tan bx + c x y period = b 180° (amplitude is undefined)
3. 3. 360° degrees = 2 π radians Radians NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART 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 π r r r r r r Radians are normally written as fractions of . π 2 π ≈ 6.28… Inaccurate
4. 4. Exact Values of Trigonometric Functions NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART 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 3 1 2 1 3 1 3 2 3 1 2 1 0 2 1 2 1 LEARN THESE
5. 5. 4 π Quadrants NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART 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 2 3 π It is useful to think of angles in terms of quadrants . 37° is in the 1 st quadrant is in the 3 rd quadrant 3 Examples π 3 4
6. 6. The Quadrant Diagram NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART y 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° x + + + +
7. 7. 7 π Quadrants and Exact Values NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART Any angle can be written as an acute angle starting from either 0° or 180° . 120° 60° sin 120° sin 60° 2 3 = 225° 45° - cos 45° = 1 2 - = cos 225° = π 6 - tan = 1 3 - tan = 6 - 6 - π 6 cos negative tan negative 7 π T + C + A + S + S + A + T + C + S + A + T + C +
8. 8. Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART Solving Trigonometric Equations Graphically NOTE It is possible to solve trigonometric equations by sketching a graph. Example Solve 2 cos x – 3 = 0 for 0 x 2 π 2 cos x = 2 3 3 cos x = x = π 6 x y 2 3 √ π 2 π π 6 or x = π 6 2 π – = 11 π 6 11 π 6 Sketching y = cos x gives: y = cos x
9. 9. Solving Trigonometric Equations using Quadrants NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART Example Solve 2 sin x + 1 = 0 for 0° x 360° 1 2 sin x = -   sin negative solutions are in the 3 rd and 4 th quadrants 45° 45° acute angle: sin ( ) = 45° x = 180° + 45° = 225° x = 360° – 45° = 315° or Trigonometric equations can also be solved algebraically using quadrants. The ‘X-Wing’ Diagram 1 2 A + C + S + T + - 1 S + A + T + C +
10. 10. 3 NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART Example 2 Solve tan 4 x + 3 = 0 for 0 x π 2 tan 4 x = 3 -   tan negative solutions are in the 2 nd and 4 th quadrants tan ( ) = π 3 4 x = π 3 π – = 2 π 3 x = π 6 or 4 x = π 3 2 π – = 5 π 3 x = 5 π 12 π π 2 3 π 2 0 (continued) Solving Trigonometric Equations using Quadrants acute angle: A + C + S + T + - 1
11. 11. Problems involving Compound Angles NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART Solve 6 sin ( 2 x + 10 ) = 3 for 0° x 360° Example sin ( 2 x + 10 ) = 2 1   solutions are in the 1 st and 2 nd quadrants 30° 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° A + C + S + T +
12. 12. NOTE Higher Maths 1 2 3 Trigonometric Functions UNIT OUTCOME SLIDE PART ( 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 S A T C   acute angle ≈ 34.8° x ≈ 180° + 34.8° ≈ 214.8° or x ≈ 360° – 34.8° ≈ 325.2° x = 90° Example Solving Quadratic Trigonometric Equations FACTORISE!