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Chapter 5 Section 5.2 Trigonometric Functions Angles in Standard Position y-axis Recall an angle in standard position is an angle that has its initial side on the positive x-axis. We can use any point on the angles terminal side to find the values of the trigonometric ratios. If the coordinates of the point P are (x,y) and the distance the point P is from the origin is r we get the following values for the trigonometric ratios. y sin r cos y x2 y2 x r x x2 y 2 y tan x x cot y In the example to the right with the coordinates of P at the point (1,3) 3 sin 10 1 cos 10 3 tan 1 1 cot 3 10 sec 1 10 csc 3 P:(x,y) r x2 y2 y x-axis x r sec x x2 y2 x r csc y x2 y 2 y P:(1,3) 3 2 r 1 r 12 32 r 10 1 2 It is important to realize that it does not matter what point you select on the terminal side of the angle the trigonometric ratios will be the same because the triangles are similar. The triangle with its vertex at P1 is similar to the triangle with its vertex at P2 and the length of the sides are proportional (equal ratios). P2 P1 Signs of Trigonometric Functions The trigonometric ratios now are defined no matter where the terminal side of the angle is. It can be in any if the four quadrants. Since the values for the xy-coordinates are different signs (±) depending on the quadrant the trigonometric ratios will be also. The value for r is always positive. The chart below shows the signs of the trigonometric ratios. x neg (-) y pos (+) x pos (+) y pos (+) x neg (-) y neg (-) x pos (+) y neg (-) Quadrant sin cos tan cot sec csc I + + + + + + II + - - - - + III - - + + - - IV - + - - + - Find the values of the six trigonometric functions for the left side of the line 2𝑥 + 3𝑦 = 0. The point (-3,2) is on the terminal side of the angle. We find the value for r (distance from the origin) first. 2 r r (3) 2 2 2 9 4 13 -3 2 3 2 3 13 13 sin cos cot tan csc sec 13 13 2 3 2 3 Any point on one side of a line can be used to determine the values of the trigonometric functions. To find the values at the 4 angles 0° , 90° , 180° , 270° , and 360° use the 4 points pictured and keep in mind 𝑟 = 1. Angle x y sin cos tan cot sec csc 90 0° 1 0 0 1 0 undefined 1 undefined 1 90° 180° 0.5 0 -1 1 0 1 0 0 -1 undefined 0 0 undefined undefined -1 1 undefined 180 1 0.5 0.5 0.5 270° 0 -1 -1 0 undefined 0 undefined -1 1 360° 1 0 0 1 0 undefined 1 undefined 270 1 0 ,360 Reciprocal Identities: 𝑦 1 sin 𝜃 = = 𝑟 csc 𝜃 𝑟 1 csc 𝜃 = = 𝑦 sin 𝜃 Quotient Identities: 𝑥 1 cos 𝜃 = = 𝑟 sec 𝜃 𝑟 1 sec 𝜃 = = 𝑥 cos 𝜃 tan 𝜃 = Pythagorean Identities: sin2 𝜃 + cos 2 𝜃 = 1 𝑦 𝑥 = sin 𝜃 cos 𝜃 𝑦 1 tan 𝜃 = = 𝑥 cot 𝜃 𝑥 1 cot 𝜃 = = 𝑦 tan 𝜃 cot 𝜃 = 1 + tan2 𝜃 = sec 2 𝜃 𝑥 𝑦 = cos 𝜃 sin 𝜃 1 + cot 2 𝜃 = csc 2 𝜃 These come from the fact that in a right triangle with sides x,y, and r we have: 𝑥 2 + 𝑦 2 = 𝑟 2 2 2 2 2 2 𝑥 𝑦 𝑥 + 𝑦 𝑟 sin2 𝜃 + cos 2 𝜃 = 2 + 2 = = 2=1 2 𝑟 𝑟 𝑟 𝑟 Trigonometric Identities can be very useful when trying to do the following problem. If 𝜃 is an angle in the second quadrant and 5 sin 𝜃 = find the other trigonometric 6 functions of t. 2 5 2 cos t 1 6 25 cos 2 t 1 36 11 2 cos t 36 11 Second quadrant cos t cosine is negative 6 5 6 11 cos t 6 sin t tan t cot t sec t csc t 5 sin t 5 5 11 6 cos t 11 11 11 6 1 11 tan t 5 1 1 6 11 cos t 11 11 6 1 6 sin t 5 Besides using Identities you can equate parts of fractions. 𝑦 − 3 sin 𝜃 = = 𝑟 2 𝑥 −1 If sec 𝜃 = −2 and the value of 𝜃 is in the cos 𝜃 = = 𝑟 2 third quadrant, find the value of the other six trigonometric functions. 𝑦 − 3 tan 𝜃 = = = 3 𝑥 −1 𝑟 2 𝑥 −1 1 3 sec 𝜃 = = 𝑥 −1 cot 𝜃 = = = = 𝑦 − 3 3 3 𝑟 = 2 𝑎𝑛𝑑 𝑥 = −1 𝑟 2 𝑥2 + 𝑦2 = 𝑟2 sec 𝜃 = = = −2 𝑥 −1 1 + 𝑦2 = 4 𝑟 2 2 3 −2 3 𝑦2 = 3 In the third csc 𝜃 = = = = quadrant y is 𝑦 − 3 −3 3 𝑦=− 3 negative.