Mrs. Volynskaya Pre-Calculus Chapter 4 Trigonometry   . Mrs. Volynskaya Pre-Calculus Chapter 4 Trigonometry   𝜽 1 1 y 𝜽 x -1 1 -1

  sin 𝜃 = cos 𝜃 = Since there is exactly one point P(x,y) for an angle 𝜃, the relations cos 𝜃 = x and sin 𝜃 = y are functions of 𝜃. Because both of these functions are defined using the unit circle, they are often called ________________. The four other trig functions can also be defined using the unit circle: tan 𝜃 = csc 𝜃 = sec 𝜃 = cot 𝜃 = 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑦 1 = 𝑥 1 circular functions 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 𝑦 𝑥 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 = 1 𝑦 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 1 𝑥 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 = 𝑥 𝑦

Example 1: Use the unit circle to find each value.   a. tan 180° tan 180° = 𝑦 𝑥 sin 180 0 = 0 sec 180 0 = -1 tan 180° = 0 −1 cos 180 0 = -1 cot 180 0 = undefined csc 180 0 = undefined tan 180° = 0 b. csc (-90°) csc (-90°) = 1 𝑦 sin −90 0 = -1 cot −90 0 = 0 csc (-90°) = 1 −1 cos −90 0 = 0 tan −90 0 = undefined csc (-90°) = -1 sec −90 0 = undefined

sin , csc cos, tan, sec, cot sin, cos, tan, sec, csc, cot none 0 1 - 1 2 3 2 90 1 2 3 2 - 2 2 2 2 2 2 2 2 120 60 - 3 2 1 2 135 45 3 2 1 2 150 30 -1 0 180 1 0 360 210 330 225 315 - 3 2 - 1 2 240 3 2 - 1 2 300 - 2 2 - 2 2 2 2 - 2 2 - 1 2 - 3 2 270 1 2 - 3 2 0 -1 tan, cot sin, cos, sec, csc cos, sec sin, tan, csc, cot

** The reference angle would be 60°. Quad. II x is neg. & y is pos. The radius of a circle is defined as a positive value. Therefore the signs of the six trig functions are determined by the signs of the coordinates of ___ and ___ in each quadrant. x y Example 2: Use the unit circle to find the values of the six trigonometric functions for a 300° angle. ** The reference angle would be 60°. Quad. II x is neg. & y is pos. Quad. I x & y are pos. ** The terminal side of the angle intersects the unit circle at a point. ** We need to find the coordinates of this point from our special triangles from Geometry. 300° Special Triangles Quad. III x & y are neg. Quad. IV x is pos. & y is neg. 30° - 60° - 90° 45° - 45° - 90° sin 60° = 𝑜𝑝𝑝. ℎ𝑦𝑝𝑜𝑡. = 𝑥 3 2𝑥 = 3 2 60° 45° 2x y 2 y x cos 60° = 𝑎𝑑𝑗. ℎ𝑦𝑝𝑜𝑡. = 𝑥 2𝑥 = 1 2 30° 45° The pt. is 1 2 , − 3 2 x 3 y

** Now that we know the point on the terminal side that intersects the unit circle, we can now find the 6 trig. functions. 1 2 ,− 3 2 = - 3 2 sin 300° = y = 1 2 cos 300° = x = − 3 2 1 2 tan 300° = 𝑦 𝑥 = - 3 csc 300° = 1 𝑦 = 1 − 3 2 = - 2 3 = - 2 3 3 sec 300° = 1 𝑥 = 1 1 2 = 2 = 1 2 − 3 2 cot 300° = 𝑥 𝑦 = - 1 3 = - 3 3

Trig Functions of an Angle in Standard Position The sine and cosine functions of an angle in standard position may also be determined using the ordered pair of any point on its ________________ and the distance between that _______ and the_______. terminal side point origin P(x, y) r y x All six trig functions can be determined using x, y, and r. (𝑟= 𝑥 2 + 𝑦 2 ) Trig Functions of an Angle in Standard Position sin 𝜃 = 𝑦 𝑟 cos 𝜃 = 𝑥 𝑟 tan 𝜃= 𝑦 𝑥 csc 𝜃 = 𝑟 𝑦 sec 𝜃= 𝑟 𝑥 cot 𝜃= 𝑥 𝑦

Example 3: Find the values of the six trigonometric functions for angle  in standard position if a point with coordinates (-6, 8) lies on its terminal side. ** 1st find r : r = (−6) 2 + 8 2 r = 100 r = 10 sin 𝜃 = 𝑦 𝑟 = 8 10 = 4 5 sec 𝜃 = 𝑟 𝑥 = 10 −6 = - 5 3 cos 𝜃 = 𝑥 𝑟 = −6 10 = - 3 5 cot 𝜃 = 𝑥 𝑦 = −6 8 = - 3 4 tan 𝜃 = 𝑦 𝑥 = 8 −6 = - 4 3 csc 𝜃 = 𝑟 𝑦 = 10 8 = 5 4

If you know the value of one of the trig functions and the quadrant in which the terminal side lies in, you can find the other five trig functions. Example 4: Suppose  is an angle in standard position whose terminal side lies in Quadrant II. If csc  = , find the values of the remaining five trigonometric functions of . 8 3 ** Since the point lies in Quad. II, x is neg. & y is pos. csc 𝜃 = 𝑟 𝑦 , so r = 8 and y = 3 sin 𝜃 = 𝑦 𝑟 = 3 8 Find x : 𝑥 2 + 𝑦 2 = 𝑟 2 cos 𝜃 = 𝑥 𝑟 = − 55 8 𝑥 2 + 9 = 64 tan 𝜃 = 𝑦 𝑥 = 3 − 55 = - 3 55 55 𝑥 2 = 55 x = ± 55 sec 𝜃 = 𝑟 𝑥 = 8 − 55 = - 8 55 55 x has to be - 55 cot 𝜃 = 𝑥 𝑦 = − 55 3