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

2. (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

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

4. (C) Triangles
To work with the 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

5. (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

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

7. (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

8. (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

9. (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

10. (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

11. (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

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