Presentation is loading. Please wait.

Presentation is loading. Please wait.

Mrs. Crespo 2011. Definition  Let θ be a non-quadrantal angle in standard position.  The reference angle for θ is the acute angle θ R that the terminal.

Similar presentations


Presentation on theme: "Mrs. Crespo 2011. Definition  Let θ be a non-quadrantal angle in standard position.  The reference angle for θ is the acute angle θ R that the terminal."— Presentation transcript:

1 Mrs. Crespo 2011

2 Definition  Let θ be a non-quadrantal angle in standard position.  The reference angle for θ is the acute angle θ R that the terminal side of θ makes with the x-axis. In simple English, a reference angle is:  non-quadrantal (does not land on the axis)  an acute positive angle (less than 90˚)  located between the terminal side and the x-axis  denoted as θ R Mrs. Crespo 2011

3 Bookmark Mrs. Crespo 2011 A reference angle is:  non-quadrantal (does not land on the axis)  an acute positive angle (less than 90˚)  located between the terminal side and the x-axis  denoted as θ R θ θ θ θRθR θRθR θRθR θRθR θRθR θRθR

4 Bookmark Examples Formulas for the reference angle θ R with 0˚ < θ < 360˚ or 0 < θ < 2 π.  Quadrant I: θ R = θ  Quadrant II: θ R = 180˚- θ or π - θ  Quadrant III: θ R = θ - 180˚ or θ - π  Quadrant IV: θ R = 360˚- θ or 2 π - θ θ = 315 ˚  Quadrant IV  θ R = 360˚- θ = 360˚- 315˚ = 45˚ Mrs. Crespo 2011 θ = -240 ˚  It is a negative angle. Add 360˚.  -240˚+ 360˚= 120˚  Quadrant II  θ R = 180˚- θ = 180˚- 120˚ = 60˚ θ = 5 π / 6  Quadrant II  θ R = π - θ = π – 5 π / 6 = 6 π / 6 – 5 π / 6 = π / 6

5 Bookmark Examples: Use reference angle to find the exact values of sin θ, cos θ, and tan θ. If θ is a non-quadrantal angle in standard position, then to find the value of a trigonometric function at θ:  Find its reference angle value θ R.  Prefix the appropriate sign. θ = 315 ˚  Quadrant IV  θ R = 360˚- θ = 360˚- 315˚ = 45˚ sin θ = y  y-values in Q IV are negative  sin 315˚ = - sin 45˚ = - √2 / 2 cos θ = x  x -values in Q IV are positive  cos 315˚ = cos 45˚ = √2 / 2 tan θ = y / x  tan 315˚= ( - √2 / 2 )/ ( √2 / 2 ) = -1 Mrs. Crespo 2011 θ = 5 π / 6  Quadrant II  θ R = π – θ = π – 5 π / 6 = 6 π / 6 – 5 π / 6 = π / 6 sin θ = y  y-values in Q II are positive  sin 5 π / 6 = + sin π / 6 = 1 / 2 cos θ = x  x -values in Q II are negative  cos 5 π / 6 = - cos π / 6 = - √3 / 2 tan θ = y / x  tan 5 π / 6 = - tan π / 6 = - √3 / 3

6 You should get. Press to enter ½ inside the parenthesis. Find an angle using calculator (TI-83). Mrs. Crespo 2011 Given sin θ = ½, find angle θ. On calculator, access sin -1 (inverse of sin) by pressing. 2ndsin 30 ⁰ 1÷2 Your Turn: Find the angle of cos θ = ½. cos -1 (½) = 60˚

7 Page 459 1-17 ODD Mrs. Crespo 2011


Download ppt "Mrs. Crespo 2011. Definition  Let θ be a non-quadrantal angle in standard position.  The reference angle for θ is the acute angle θ R that the terminal."

Similar presentations


Ads by Google