1 / 7

3.4 Circular Functions

3.4 Circular Functions. x 2 + y 2 = 1 is a circle centered at the origin with radius 1 call it “The Unit Circle”. Any point on this circle can be defined in terms of sine & cosine ( x , y )  ( cos θ , sin θ ). (1, 0).

marlee
Download Presentation

3.4 Circular Functions

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. 3.4 Circular Functions

  2. x2 + y2 = 1 is a circle centered at the origin with radius 1 call it “The Unit Circle” Any point on this circle can be defined in terms of sine & cosine (x, y)  (cosθ, sin θ) (1, 0) Ex 1) For the radian measure , find the value of sine & cosine. (0, 1)

  3. We can draw a “reference” triangle by tracing the x-value first & then the y-value to get to a point. Use Pythagorean Theorem to find hypotenuse. (x, y) r y θ x Ex 2) The terminal side of an angle θ in standard position passes through (3, 7). Draw reference triangle & find exact value of cosθ and sin θ. (3, 7) r 7 θ 3

  4. Ex 3) Find the exact values of cosθ and sin θ for θ in standard position with the given point on its terminal side. –1 θ r

  5. cosθ = 0 Reminder: (–, +) (+, +) I II sinθ = 0 sinθ = 0 III IV (–, –) (+, –) cosθ = 0 Ex 4) State whether each value is positive, negative, or zero. a) cos 75° b) sin (–100°) c) positive negative zero

  6. Ex 5) An angle θis in standard position with its terminal side in the 2nd quadrant. Find the exact value of cosθif 10 8 Pythag says: x2 + 82 = 102 x2 = 36 x = ±6 x –6 why negative? so…

  7. Homework #304 Pg 145 #1–49 odd

More Related