2 Our method of using right triangles only works for acute 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° =135°reference angleIf 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.What is the measure of this reference angle?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.
3 Let's label the sides of the triangle according to a 45-45-90 triangle Let's label the sides of the triangle according to a triangle. (The sides might be multiples of these lengths but looking as a ratio that won't matter so will work)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 =135°145°-1Now we are ready to find the 6 trig functions of 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.
4 Notice the -1 instead of 1 since the terminal side of the angle is in quadrant II where x values are negative. =135°145°-1We 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.
5 Let's use this idea to find the 6 trig functions for 210° 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=210°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?30°-12210°-180°=30°Label the sides of the triangle and include any negative signs depending on if x or y values are negative in the quadrant.The reference angle is the amount past 180° of
6 You will never put a negative on the hypotenuse 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.210°30°-12You should be thinking csc is the reciprocal of sin and sin is opposite over hypotenuse so csc is hypotenuse over opposite.
7 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 picture5Now drop a perpendicular line from the terminal side to the x-axis-1213(5, -12)Label the sides of the triangle including any negatives. You know the two legs because they are the x and y values of the pointUse the Pythagorean theorem to find the hypotenuse
8 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.5We'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.-1213(5, -12)
9 All trig functions positive In quadrant I both the x and y values are positive so all trig functions will be positive+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 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.
10 _ _ + _ In quadrant III, x is negative and y is 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 IV, x is positive and y is negative .So any functions using opposite will be negative.sin is - cos is + tan is -
11 All trig functions positive To help remember these sign we look at what trig functions are positive in each quadrant.sin is + cos is - tan is -All trig functions positiveSATCsin is - cos is - tan is +sin is - cos is + tan is -Here is a mnemonic to help you remember. (start in Quad I and go counterclockwise)StudentsAllTakeCalculus
12 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.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).Try this with 90°(0, 1)dividing by 0 is undefined so the tangent of 90° is undefined
13 Remember x is adjacent, y is opposite and hypotenuse here is 1 Let's find the trig functions of (-1, 0)Remember x is adjacent, y is opposite and hypotenuse here is 1
14 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.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.62°-298°422°
15 AcknowledgementI wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum.Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramar