• Like
Trigonometric Function of General Angles Lecture
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

Trigonometric Function of General Angles Lecture

  • 1,014 views
Published

Lesson Objectives …

Lesson Objectives

Trigonometric Functions of Angles
Trigonometric Function Values
Could find the Six Trigonometric Functions
Learn the signs of functions in different Quadrants
Could easily determine the signs of each Trigonometric Functions
Solve problems involving Quadrantal Angles
Find Coterminal Angles
Learn to solve using reference angle
Solve problems involving Trigonometric Functions of Common Angles
Solve problems involving Trigonometric Functions of Uncommon Angles

Published in Business , Technology
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
No Downloads

Views

Total Views
1,014
On SlideShare
0
From Embeds
0
Number of Embeds
1

Actions

Shares
Downloads
41
Comments
0
Likes
2

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. www.FroydWess.com Presents: Trigonometric Function of General Angles credit: Shawna Haider
  • 2. OF GENERAL ANGLES
  • 3. Our method of using right triangles only works for acute angles. Now we will see how we can find the trig function values of any angle. To do this we'll place angles on a rectangular coordinate system with the initial side on the positive x-axis.  HINT: Since it is 360 all the way around a circle, half way around (a straight line) is 180 If  is 135 , we can find the angle formed by the negative x-axis and the terminal side of the angle. This is an acute angle and is called the reference angle. reference angle What is the measure of this reference angle?  =135 180 - 135 = 45 Let's make a right triangle by drawing a line perpendicular to the x-axis joining the terminal side of the angle and the x-axis.
  • 4. Let's label the sides of the triangle according to a 45-45-90 triangle. (The sides might be multiples of these lengths but looking as a ratio that won't matter so will work) 45  =135 The values of the trig functions of angles and their reference angles are the same except possibly they may differ by a negative sign. Putting the negative on the 1 will take care of this problem. -1 1 2 211  Now we are ready to find the 6 trig functions of 135 This is a Quadrant II angle. When you label the sides if you include any signs on them thinking of x & y in that quadrant, it will keep the signs straight on the trig functions. x values are negative in quadrant II so put a negative on the 1
  • 5. -1 45  =1351 2 1 2 sin135 22 o h     Notice the -1 instead of 1 since the terminal side of the angle is in quadrant II where x values are negative. 1 1 10 135tan    a We are going to use this method to find angles that are non acute, finding an acute reference angle, making a triangle and seeing which quadrant we are in to help with the signs. 2 2 2 1 135cos     h a
  • 6. Let  denote a nonacute angle that lies in a quadrant. The acute angle formed by the terminal side of  and either the positive x-axis or the negative x-axis is called the reference angle for . Let's use this idea to find the 6 trig functions for 210 First draw a picture and label  (We know that 210 will be in Quadrant III) Now drop a perpendicular line from the terminal side of the angle to the x-axis The reference angle will be the angle formed by the terminal side of the angle and the x-axis. Can you figure out it's measure? =210 210 -180 =30 The reference angle is the amount past 180 of  30 Label the sides of the 30-60-90 triangle and include any negative signs depending on if x or y values are negative in the quadrant. 2 -1 3
  • 7. 30 210 2 -1 3 You will never put a negative on the hypotenuse. Sides of triangles are not negative but we put the negative sign there to get the signs correct on the trig functions. 210csc You should be thinking csc is the reciprocal of sin and sin is opposite over hypotenuse so csc is hypotenuse over opposite. 2 1 2   210tan 3 3 3 1    210cos 2 3
  • 8. Using this same triangle idea, if we are given a point on the terminal side of a triangle we can figure out the 6 trig functions of the angle. Given that the point (5, -12) is on the terminal side of an angle , find the exact value of each of the 6 trig functions. First draw a picture (5, -12) Now drop a perpendicular line from the terminal side to the x-axis Label the sides of the triangle including any negatives. You know the two legs because they are the x and y values of the point 5 -12 Use the Pythagorean theorem to find the hypotenuse     222 125 h 13h 13
  • 9. Given that the point (5, -12) is on the terminal side of an angle , find the exact value of each of the 6 trig functions. (5, -12) 5 -12 13       cottan seccos cscsin   We'll call the reference angle . The trig functions of  are the same as  except they possibly have a negative sign. Labeling the sides of triangles with negatives takes care of this problem. 12 13 o h   13 5  h a 12 5 o a   12 13   o h 5 13  a h 12 5   o a
  • 10. The Signs of Trigonometric Functions Since the radius is always positive (r > 0), the signs of the trig functions are dependent upon the signs of x and y. Therefore, we can determine the sign of the functions by knowing the quadrant in which the terminal side of the angle lies.
  • 11. In quadrant I both the x and y values are positive so all trig functions will be positive + + All trig functions positive In quadrant II x is negative and y is positive. _ +  We can see from this that any value that requires the adjacent side will then have a negative sign on it. Let's look at the signs of sine, cosine and tangent in the other quadrants. Reciprocal functions will have the same sign as the original since "flipping" a fraction over doesn't change its sign. sin is + cos is - tan is -
  • 12. _ _  In quadrant IV, x is positive and y is negative . _ +  So any functions using opposite will be negative. Hypotenuse is always positive so if we have either adjacent or opposite with hypotenuse we'll get a negative. If we have both opposite and adjacent the negatives will cancel sin is - cos is + tan is - In quadrant III, x is negative and y is negative. sin is - cos is - tan is +
  • 13. All trig functions positive sin is + cos is - tan is - sin is - cos is + tan is - sin is - cos is - tan is + To help remember these sign we look at what trig functions are positive in each quadrant. AS T C Here is a mnemonic to help you remember. (start in Quad I and go counterclockwise) AllStudents Take Calculus
  • 14. To find the sine, cosine, tangent, etc. of angles whose terminal side falls on one of the axes , we will use the unit circle. 3 (..., , , 0, , , , 2 ,...) 2 2 2        (1, 0) (0, 1) (-1, 0) (0, -1) 0 2  3 2   Unit Circle:  Center (0, 0)  radius = 1  x2 + y2 = 1 Trigonometric Functions Quadrantal Angle
  • 15. What about quadrantal angles? We can take a point on the terminal side of quadrantal angles and use the x and y values as adjacent and opposite respectively. We use the x or y value that is not zero as the hypotenuse as well. Try this with 90 (0, 1) We can take a point on the terminal side of quadrantal angles and use the x and y values as adjacent and opposite respectively. We use the x or y value that is not zero as the hypotenuse as well (but never with a negative). 90sin  h o 1 1 1  90cos  h a 0 1 0  90tan  a o 0 1 dividing by 0 is undefined so the tangent of 90 is undefined 90cosec 1 1 1  90sec undef 0 1 90cot 0 1 0 
  • 16. Let's find the trig functions of  (-1, 0) sin  h o 0 1 0  cos  h a 1 1 1   tan  a o 0 1 0  cosec undef 0 1 sec 1 1 1   cot undef 0 1 Remember x is adjacent, y is opposite and hypotenuse here is 1
  • 17. Coterminal angles are angles that have the same terminal side. 62 , 422 and -298 are all coterminal because graphed, they'd all look the same and have the same terminal side. 62 422 -298 Since the terminal side is the same, all of the trig functions would be the same so it's easiest to convert to the smallest positive coterminal angle and compute trig functions.
  • 18. Reference Angles for The reference angles in Quadrants II, III, and IV.  ′ =  –  (radians)  ′ = 180 –  (degrees)  ′ =  –  (radians)  ′ =  – 180 (degrees)  ′ = 2 –  (radians)  ′ = 360 –  (degrees)
  • 19. Finding Reference Angles angle ′.Find the reference a.  = 300 b.  = 2.3 c.  = –135
  • 20. Trigonometric Functions of Real Numbers
  • 21. Trigonometric Functions of Any Angle Definitions of Trigonometric Functions of Any Angle Let  be an angle in standard position with (x, y) a point on the terminal side of  and 2 2 r x y  sin csc cos sec tan cot y r r y x r r x y x x y             y x  (x, y) r
  • 22. Evaluate each trigonometric function using Reference Angle. a. cos b. tan(–210) c. csc Evaluating Trigonometric Functions
  • 23. (a) – Solution Because  = 4/3 lies is in Quadrant III, the reference angle As shown in Figure (a). Moreover, the cosine is negative Quadrant III, so in Figure (a)
  • 24. (b) – Solution Because –210 + 360 = 150, it follows that –210 is 150.coterminal with the second-quadrant angle is ′ = 180 – 150 = 30,So, the reference angle as shown in Figure (b) Figure (b)
  • 25. cont’d Finally, because the tangent is have negative in Quadrant II, you tan(–210) = = (–) tan 30 . (b) – Solution
  • 26. (c) – Solution Because (11/4) – 2 = 3/4, it follows that 11/4 is coterminal with the second-quadrant angle 3/4. –is ′ =  (3/4) /4,So, the reference angle = as shown in Figure (c) Figure (c)
  • 27. cont’d Because the cosecant is positive in Quadrant II, you have (c) – Solution
  • 28. Evaluating Trigonometric Functions of .Let (–3, 4) be a point on the terminal . side Find the sine, cosine, and tangent of Solution: x = –3, y = 4,
  • 29. Solution
  • 30. Try this: 13 -5 y x  (-12, -5) -12 sin cos tan csc sec cot y r x r y x r y r x x y                  
  • 31. Given and , find the values of the five other trig function of . Next Problems... 8 cos 17    cot 0  Given and , find the values of the five other trig functions of . 3 cot 8    2   
  • 32. Find the value of the six trig functions for For Quadrantal Angles 2     sin 2 cos 2 tan 2 1 csc 2 1 sec 2 cot 2 y x y x y x x y                                                       (1, 0) (0, 1) (-1, 0) (0, -1) 0 3 2   2    
  • 33. Find the value of the six trig functions for For Quadrantal Angles 7              sin 7 cos 7 tan 7 1 csc 7 1 sec 7 cot 7 y x y x y x x y                  
  • 34. Find one positive and one negative coterminal angle of For Coterminal Angles 3 4    
  • 35. Using Coterminal Angle to Find the Exact Value of a Trigonometric Funcion
  • 36. Using reference angles and the special reference triangles, we can find the exact values of the common angles. To find the value of a trig function for any common angle  1. Determine the quadrant in which the angle lies. 2. Determine the reference angle. 3. Use one of the special triangles to determine the function value for the reference angle. 4. Depending upon the quadrant in which  lies, use the appropriate sign (+ or –). Trig Functions of Common Angles
  • 37. Examples Give the exact value of the trig function (without using a calculator). 1. 2. 5 sin 6  3 cos 4      
  • 38. Trig Functions of “Uncommon” Angles To find the value of the trig functions of angles that do NOT reference 30 , 45 , or 60 , and are not quadrantal, we will use the calculator. Round your answer to 4 decimal places, if necessary.  Make sure the Mode setting is set to the correct form of the angle: Radian or Degree  To find the trig functions of csc, sec, and cot, use the reciprocal identities.
  • 39. Examples Evaluate the trig functions to four decimal places. 1. 2.sec( 2.5)  csc 23 38' 45"
  • 40. Online Notes and Presentations wwww.FroydWess.com Visit: