|
Special Angles for Trigonometric Functions
Motivation: Previously, we have seen the definition of trigonometric functions associated with right triangles. And once an angle and the length of a side are known, we can find the lengths of the rest of sides in that right triangle with the help of trigonometric functions. And similarity theorem in classical geometry indicates that it is meaningful to do that because the relationship will not be changed when you use “larger” or “smaller” triangles.
But how do we know the values of trigonometric functions associated with different angles? The systematic approach to find the values of trigonometric functions can be found in Calculus – known as Taylor’s formula. But for some special angles, we can easily find the values for the corresponding trigonometric functions from classical geometry. We start with the result from Equilateral Triangle.
Example ( Equilateral Triangle ) ABC is an equilateral triangle such that the length of each side is 2. If , then we have B = 600 , BAM = 300 , =1, = .
So, sin600 = = cos600 = = sin300 = = cos300 = = tan600 = tan300 =
Example ( Isosceles Triangle with Right Angle ) B=900 , = =1
= ( via Pythagorean Theorem )
With the property of isosceles, we know that C= A. And the sum of interior angles of triangle is 1800 and B=900, we have C= A=450
So, sin450 = = = cos450 = tan450 = =1
Example: B=900 , =1, =2, =2 (1) Find the angle of D . (2) Find the value of tan( D) . Sol: (1) = = Thus, C = 600 . Furthermore, = C = 2 D D = 300 (2) tan300 = = =
From the examples above, we can find the value of tan300 from different geometry figures and all the results are the same. Recall what we have mentioned at the beginning of this section : this is the result of similarity theorem.
Example ( sin150 ): B=900 , =1, =2, =2 Use the diagram to find sin150 .
Sol: 1. = = C = 300 2. = C = 2 D D = 150 3. 2 = 2 + 2 = ( 2 + )2 + 1 = 8 + 4 = 8 + 2 = 6 + 2 + 2 = ( + )2 So, = +
4. sin150 = = =
And it is natural to find tan150 = = 2- cos150 = = =
Example ( sin22.50 ) B=900 , = =1, = . Use the diagram to find sin22.50 .
Sol: 1. With similar approach as shown in the previous example, D = 22.50 = = 2. 2 = 2 + 2 = ( )2 + 1 = 4 + 2 =
3. sin22.50 =
Example ( sin180 ) As shown below, = = =1 , and = . Use the diagram to shown that sin180 =
Proof:
Let’s denote the length of as x.
1. = C = ABC 2. = BDC = 2 A 3. = C = BDC = 2 A
Inside ABC, we know that A + C + ABC = 1800 A = 360 , C=720, ABC=720, BDC=720 So, DBC=360 .
4. ABC BCD ; in other words,
x2 + x -1 = 0 x = But the length can not be negative. Thus =
5. From B, construct a line perpendicular to as shown. = is the bisector of DBC; is also the perpendicular bisector of . sin180 = = = |
