Trigonometry functions of general angles reference anglesPresentation Transcript
Now lets look at the special angles called the quadrantal angles. The quadrantal angles are those angles that lie on the axis of the Cartesian coordinate system: , , , and .
We also need to be able to recognize these angles when they are given to us in radian measure. Look at the smallest possible positive angle in standard position, other than 0 , yet having the same terminal side as 0 . This is a 360 angle which is equivalent to . radians If we look at half of that angle, we have Looking at the angle half-way between 180 and 360 , we have 270 or radians which is of the total (360 or ). radians Moving all the way around from 0 to 360 completes the circle and and gives the 360 angle which is equal to radians. Looking at the angle half-way between 0 and 180 or , we have 90 or .
Add the equivalent degree measure to each of these quadrantal angles. We can count the quadrantal angles in terms of . 0 radians Notice that after counting these angles based on portions of the full circle, two of these angles reduce to radians with which we are familiar, . radians We can approximate the radian measure of each angle to two decimal places. One of them, you already know, . It will probably be a good idea to memorize the others. Knowing all of these numbers allows you to quickly identify the location of any angle.
We can find the trigonometric functions of the quadrantal angles using this definition. We will begin with the point (1, 0) on the x axis. As this line falls on top of the x axis, we can see that the length of r is 1. Remember the six trigonometric functions defined using a point ( x , y ) on the terminal side of an angle, . (1, 0) 0 radians radians or For the angle 0 , we can see that x = 1 and y = 0. To visualize the length of r, think about the line of a 1 angle getting closer and closer to 0 at the point (1, 0).
It will be just as easy to find the trig functions of the remaining quadrantal angles using the point (x, y) and the r value of 1. 0 radians radians (1, 0) or Using the values, x = 1, y = 0, and r = 1, we list the six trig functions of 0. And of course, these values also apply to 0 radians, 360 , 2 radians, etc.
TRIGONOMETRY FUNCTIONS OF GENERAL ANGLES
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.
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 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
-1 45 ° =135 ° 1 Notice the -1 instead of 1 since the terminal side of the angle is in quadrant II where x values are negative. 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.
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
30 ° 210 ° 2 -1 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. You should be thinking csc is the reciprocal of sin and sin is opposite over hypotenuse so csc is hypotenuse over opposite.
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 13
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 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.
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 - _ +
_ _ 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 + _ +
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. A S T C Here is a mnemonic to help you remember. (start in Quad I and go counterclockwise) A ll S tudents T ake C alculus
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). dividing by 0 is undefined so the tangent of 90 ° is undefined
Let's find the trig functions of (-1, 0) Remember x is adjacent, y is opposite and hypotenuse here is 1
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.
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au