TF.02.2 - Trigonometric Ratios of Special Angles MCR3U - Santowski

(A) 45-45-90 Triangles If we have an isosceles triangle and make each non-hypotenuse side 1 unit long, then we have the hypotenuse as 2 Each of our “base” angles are then 45 degrees Then sin(45) = 1/ 2 = 2 / 2 Then cos(45)= 1/ 2 = 2 / 2 Then tan(45)= 1/1 = 1

(B) 45-45-90 Triangles on the Cartesian Plane we can now put our 45-45-90 triangle into the Cartesian plane and investigate the trigonometric ratios of other key angles like 135, 225, 315, etc..

(C) 30-60-90 Triangles To work with the 30-60-90 triangle, we will start with an equilateral triangle, where each angle is 60 and we will set each side to be 2 units long. From one vertex, we will simply drop an altitude (to create a right angle) to the opposite side, thereby bisecting the opposite side into two equal lengths of 1 unit The altitude has also bisected the top angle (so we now have a 30 degree angle) So the altitude that we just drew measures (22 – 12) or 3 unit So now the ratios can be determined

(D) Ratios in the 30-60-90 Triangle cos 30 = 3/2 tan 30 = 1/ 3 = 3/3 sin 60 = 3/2 cos 60 = ½ tan 60 = 3/1 = 3

(E) 30-60-90 Triangles on the Cartesian Plane we can now put our 30-60-90 triangle into the Cartesian plane and investigate the trigonometric ratios of other key angles like 120, 150, 210, 240, 300, 330 etc..

(F) Trig Ratios of Angles of Multiples of 90 To understand the trig ratios of angles of 90, 180, 270, 360, etc, we will simply go back to the Cartesian plane and work with angles from ordered pairs on the Cartesian plane 90 degrees  let’s put a point at (0,3) 180 degrees let’s put a point at (-2,0) 270 degrees  point at (0,-4) 360 degrees (or 0 degrees)  point at (1,0) And recall that sin(A) = y/r, cos(A) = x/r and tan(A) = y/x

(F) Trig Ratios of Angles of Multiples of 90 sin 0 = 0/r = 0 cos 0 = r/r = 1 tan 0 = 0/r = 0 sin 90 = 1 cos 90 = 0 tan 90 = undefined sin 180 = 0 cos 180 = -1 tan 180 = 0 sin 270 = -1 cos 270 = 0 tan 270 = undefined sin 360 = sin 0 = 0 cos 360 = cos 0 = 1 tan 360 = tan 0 = 0

(G) Examples Ex 1. Determine the sine of 495° (i) first we determine what quadrant the angle lies in  495°-360° = 135° which is thus a second quadrant angle (ii) then subtract 180° - 135° = 45°, so we have a 45° angle in the second quadrant (iii) now, simply recall sin(45°) = 1/2 (iv) now account for the quadrant, as the sine ratio is positive in the second quadrant  so the final answer is +1/ 2

(G) Examples Evaluate cos(-150°) (i) first we determine what quadrant the angle lies in  360°+(-150°) = 210° which is thus a third quadrant angle (ii) then subtract 210° - 180° = 30°, so we have a 30° angle in the second quadrant (iii) now, simply recall cos(30°) = 3/2 (iv) now account for the quadrant, as the cosine ratio is negative in the third quadrant  so the final answer is - 3/2

(H) Internet Links Now try a couple of on-line “quizzes” to see how well you understand the trig. ratios of standard special angles: Introductory Exercises from U. of Sask EMR Moderate Exercises from U. of Sask EMR Advanced Exercises from U. of Sask EMR

(I) Classwork & Homework Nelson text, Page 532, Q1-4,6,8-12eol,