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Section 4 - Trig Functions of Any Angle

Page history last edited by PBworks 16 years, 3 months ago

 

4.4 --> Trigonometric Functions of Any Angle

 

 

 

Positive Trigonometric Values

 

Sine &

Cosecant

 All

 

Tangent & Cotangent

   Cosine & Secant

 

"All Students Take Calculus"

 

All are positive in Quadrant I

Sine and Cosecant are positive in Quadrant II

Tangent and Cotangent are positive in Quadrant III

Cosine and Secant are positive in Quadrant IV

 

 

Ex) Point (1, eq=\sqrt{3}) is on the terminal side of an angle. SIN > 0, COS > 0.  Find the six trigonometric ratios and name the quadrant.

 

 

 

Both SIN and COS are positive, so the triangle should be drawn in...Quadrant I

 

 

sin(eq=\theta)=eq=\frac{\sqrt{3}}{2}         cos(eq=\theta)=1/2            tan(eq=\theta)=eq=\sqrt{3}           csc(eq=\theta)= eq=\frac{2\sqrt{3}}{3}            sec(eq=\theta)= 2            cot(eq=\theta)= eq=\frac{\sqrt{3}}{3}

 

 

 

 

Reference Angle - Angle formed between the terminal side of an angle and the X-Axis

 

  • Always Acute
  • Always Positive
  • Alpha (eq=\alpha) or Theta Prime (eq=\theta')

 

 

 

 Ex) Find Reference Angle

 

    A. eq=\theta = 186°

       - Nearest X-Axis = 180°

       - eq=\alpha = 6°

                                  

                                                   

    B. eq=\theta = 91°

       - Nearest X-Axis = 180°

       - eq=\alpha = 89°

 

                                                 

 

    C. eq=\theta = 285°

       - Nearest X-Axis = 0°

       - eq=\alpha =75°

 

 

                                                     

Exact Values...

 

Ex) Cos 150 = ?

 

  Step 1: Find the Refence Angle             Nearest X-Axis = 180°, eq=\alpha = 30°

  Step 2: Cos 30° = eq=\frac{\sqrt{3}}{2}

 

 

 

 

Ex) Cos 0° = ?

 

Step 1 --> Find point on Unit Circle             -------->            Point at (1,0)

Step 2 --> Find 6 Trigonometric values

 

 

Sin eq=\theta = Y/R     --> 0

Cos eq=\theta = X/R    --> 1

Tan eq=\theta = Y/X    --> 0

Csc eq=\theta = R/Y    --> 0

Sec eq=\theta = X/R    --> 1

Cot eq=\theta = X/Y    --> 0

 

 

 

Ex) Find two values of eq=\theta that satisfy Sin eq=\theta = .6

 

Step 1 --> Find eq=\theta                                    ArcSin .6 = 36.87°

Step 2 --> Pick Quadrants                     Sin Positive in Quadrants I & II

Step 3 --> Subtract value from 180°      Values = 36.87°, 134.13°

 

 

 

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Woot!

 

Comments (1)

Anonymous said

at 8:37 pm on Jan 8, 2008

Looks great! 20/20

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