Download Exact values for trigonometric ratios

MTH 250 Graded Assignment 9, part two
Exact values for trigonometric ratios
The 30  60  90 and 45  45  90 triangles are familiar tools used to work out the exact
values of the trig ratios for those angles (the idea being that, instead of memorizing all the ratios
for 30, 60, and 45 , you memorize the proportions on the triangles, do a quick sketch, and
then work out the ratios from there using the definitions).
The goal here is to derive/prove that those lengths associated with the angles are correct.
45 45 90
Start by constructing a right triangle on a 45 angle, with side length 1. You can
take the construction as a given; we won't justify all the steps that go into proving
it's constructable (this is essentially segment construction to measure out side
length, existence of a perpendicular, and three noncollinear points determining a
triangle).
By the Euclidean angle sum theorem, the value of
B must be [Fill in 1].
Since [Fill in 2 – what angles are congruent?] this is an [Fill in 3] triangle with base AB . [Fill in
4 – justification].
Since the triangle is [Fill in 3 again], we know that [Fill in 5 – sides?] by [Fill in 6 – justification],
and therefore BC  1 .
Finally, can use the [Fill in 7 – theorem] to calculate the length AC  2 .
This (plus the definitions of the trig ratios) allows us to calculate the following values:
sin 45 
cos 45 
csc 45 
sec 45 
tan 45 
cot 45 
30 60 90
Start by constructing an equilateral triangle of side length 2. In Euclidean
geometry, equilateral triangles are equiangular, so the values of all the
angles are 60 .
Construct an altitude (by definition, this is perpendicular to the base).
[1] Write a short proof that
BAD  CAD :
[2] Use this to prove that AD bisects both the segment BC and the angle
BAC .
Therefore, BD  [Fill in 3] and m BAD  [Fill in 4].
The remaining length AD can be computed using the [Fill in 5] [So go ahead and do that.
AD  [Fill in 6]]
Note that the values of the trigonometric ratios for 30 and 60
were already computed in the notes, using the proportions from
the 30  60  90 triangle, so you don't have to do that bit
again. (Be sure you know them, obviously!)