1 / 12

(1,0)

(0,1). ( –1,0). (1,0). (0,–1). Note 2: The Unit Circle. The unit circle has a radius of 1 with a centre of (0,0). P( x , y ). . (0,1). Let ’ s consider a blow-up of an angle in the first quadrant.

edan-velazquez
Télécharger la présentation

(1,0)

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. (0,1) ( –1,0) (1,0) (0,–1) Note 2: The Unit Circle The unit circle has a radius of 1 with a centre of (0,0) P(x, y) 

  2. (0,1) Let’s consider a blow-up of an angle in the first quadrant The radius from this angle  cuts the circle at point P whose coordinates are (x, y) 1 P(x,y) 1 y Label the points O and R, and drop perpendicular from P to the x-axis to make  OPR  R O x (1,0)

  3. (0,1) 1 P(x,y) 1 x = cos  y y = sin   R O x (1,0) Using the Triangle OPR, (cos, sin ) Therefore, the co-ordinates of P are..

  4. (0,1) 1 P(x,y) 1 y  R O x (1,0) By Pythagoras x2 + y2 = 12 So: cos2 + sin2 = 1

  5. We can use the unit circle to find values for cosΘ and sinΘ without the using the calculator Example: Use the Unit Circle to find the value of: sin 30° cos 45° Check on your calculator Sin 30° = 0.5 Cos 45° = 0.7071 ≈ 0.5 ≈ 0.7

  6. Exercise 11A Page 228

  7. STARTER: Use the Unit Circle to find the value of: sin 60° cos 72° sin 23° cos 36° ≈ 0.85 ≈ 0.3 ≈ 0.4 ≈ 0.8

  8. (0,1) ( –1,0) (1,0) (0,–1) This unit circle has Four QUADRANTS Q2 Q1  Q3 Q4 All angles are measured ANTICLOCKWISE from the positive x-axis

  9. x = cos y = sin  (0,1) ( –1,0) (1,0) (0,–1) Q1 angles.... x > 0 y > 0 As both x and y are positive cos > 0 sin  > 0

  10. x = cos y = sin  (0,1) x < 0 y > 0 ( –1,0) (1,0) (0,–1) Q2 angles.... As x is negative and y is positive cos < 0 sin  > 0 These are all the obtuse angles between 90 & 180. You may have noticed their cosines were negative

  11. 150° 30° Note 3: Obtuse Angles - Quadrant 2 Consider the points P and Q P = (cos 30, sin 30) Q = (cos 150, sin 150) Their y-coordinates are the same, so Q P sin 150 = sin 30 Their x-coordinates are opposite, so cos 150 = – cos 30 This applies to all Q1 & 2 angles related this way: cos (180 – ) = – cos  Formulae for angles in Quadrant 2 sin (180 – ) = sin 

  12. Exercise 11B Page 229

More Related