Special Angles in Trigonometry
and the Calculation of Their Trig Ratios:
Angles 0, 30°, 45°, 60°, and 90° are usual angles that people have a tendency to often use them in designs. For this reason, it is helpful to have the values of sine, cosine, tangent, and cotangent of these angles memorized for prompt use. You will find a very logical way to do this in the following. You will also see how easy it is to calculate these from simple geometry.
sin(30°) = ?
It is easy to prove that in any right triangle with a 30° angle in it, the side opposite to that 30° angle is half of the hypotenuse.
already, then C=60°.
A is the right angle in which we select
a 30° angle by line
segment AM deliberately. What is left of
A is therefore 60°.
Triangle AMB becomes isosceles in which
MB = MA. (1)
Triangle AMC becomes equilateral for which
MC = CA = MA (2)
Comparing (1) with (2), we get:
MB = MC = CA, or
CA = (1/2)BC, or, CA/BC = 1/2.
cos(30°) = ?, tan(30°) = ?, cot(30°) = ?
Example 1: In the above Table, left column, it is found that cos30° = √3/2. Since sin30° = 1/2, we may use sin2(30°) + cos2(30°) = 1 to solve for cos30°.
Solution: Substituting for sin(30°), we get:
(1/2)2 + cos2(30°) =1, or, cos2(30°) = 3/4, or, cos(30°) =√3/2.
Example 2: In the above Table, middle column, it is found that tan30° =√3/3.
Since cos30°=√3/2, we may use 1+tan2(30°) = 1/cos2(30°) to solve for tan30°.
Solution: Substituting for cos(30°), we get:
1 + tan2(30°) =1/(√3/2)2, or, tan2(30°) = 4/3 - 1,
or, tan2(30°) = 1/3, or, tan(30°) = 1/√3,
or after rationalization, tan(30°) = √3/3 as expected.
Example 3: In the above Table, right column, it is found that cot30° =√3.
Since tan30°=√3/3, we may use cot(30°) = 1/tan(30°) to solve for cot30°.
Solution: Substituting for tan(30°), we get:
cot(30°) =1/(√3/3), or, cot(30°) = 3/√3, or, cot(30°) =√3 as expected.
Example 4: Calculate the trig. ratios for the 120° angle.
Solution: 120° ends in the 2nd Quad. where only sine is positive.
(a) sin120° = ? sin120° = sin(90°+30°) = +cos(30°) = +√3/2.
(b) cos120° = ? cos120° = cos(π/2+30°) = -sin(30°) = -1/2.
(c) tan120° = ? tan120° = tan(π/2+30°) = -cot(30°) = -√3.
(d) cot120° = ? cot120° = cot(π/2+30°) = -tan(30°) = -√3/3.
Example 5: Calculate the trig. ratios for the 210° angle.
Solution: 210° ends in the 3rd Quad. where only tangent and cotangent are positive.
(a) sin210° = ? sin210° = sin(180°+30°) = -sin(30°) = -1/2.
(b) cos210° = ? cos210° = cos(π+30°) = -cos(30°) = -√3/2.
(c) tan210° = ? tan210° = tan(π+30°) = +tan(30°) = +√3/3.
(d) cot210° = ? cot210° = cot(π+30°) = +cot(30°) = +√3.
sin(45°) = ?
It is easy to see that any right triangle that has a 45° angle is necessarily isosceles because the other acute angle in it becomes 45° as well. In any triangle with two equal angles, the sides opposite to those angles are also equal. The following it the method to calculate sine, cosine, tangent, and cotangent of 45°:
Example 6: Given sin45° = √2/2, calculate other trig. ratios of 45° angle.
Solution: 45° angle ends in the 1st Quad. where all trig. ratios are are positive.
= ? From the proof above:
sin(45°) = √2/2.
(b) cos(45°) = ?
cos2(45°) + sin2(45°) = 1, or
cos2(45°) = 1- sin2(45°), or,
cos2(45°) = 1 - (√2/2)2. or,
cos2(45°) = 1 - 1/2 = 1/2 or,
cos(45°) = +√2/2. (as well)
= (√2/2)/(√2/2), or
tan (45°) = 1.
cot45° = 1/tan45° = 1/1.
cot (45°) = 1.
Summary of 45° Trig. Ratios:
|sin 45° = √2/2||cos 45° = √2/2|
|tan 45° = 1||cot 45° = 1|
The Trig. Ratios of 60° Angle
Example 7: Use the trig. ratios of 30° angle to calculate the trig. ratios of 60° angle.
sin(π/2 - x) = cos(x),
cos(π/2 - x) = sin(x),
tan(π/2 - x) = cot(x), and
cot(π/2 - x) = tan (x),
if we let x = 30° , we get:
cos60° = sin 30° = 1/2
tan60° = cot30° = √3, and
cot60° = tan30° = √3/3.
Trig Ratios of 0°, 30°, 45°, 60°, and 90°Angles
The following chart gives you a very easy way to remember the important ratios:
Example 8: Find the trig. ratios of 315° angle.
Solution: 315° angle ends in the 4th Quadrant where only cosine is positive.
sin(315°) = sin(2π - 45°) = -sin45° = -√2/2.
cos(315°) = cos(2π - 45°) = +cos45° = +√2/2.
tan(315°) = tan(2π - 45°) = -tan45° = -1.
cot(315°) = cot(2π - 45°) = -cot45° = -1.
Example 9: Find the trig. ratios of 330° angle.
Solution: 330° angle ends in the 4th Quadrant where only cosine is positive.
sin(330°) = sin(2π - 30°) = -sin30° = -1/2.
cos(330°) = cos(2π - 30°) = +cos30° = +√3/2.
tan(330°) = tan(2π - 30°) = -tan30° = -√3/3.
cot(330°) = cot(2π - 30°) = -cot30° = -√3.
Find the trig. ratios for each of the following angles:
1) 210°, 315°, 120°, 4π/3, 11π/6, 3π/4.
2) Calculate the following expression:
3) Verify the following equalities:
4) sin(π -a)sin(2π +a) + cos(π/2 -a)cos(π/2 +a) + cos2 (π + a) = 1/(1 + tan2a).
5) tan(π + x)cot(x - π) - cos(2π - x)cos(2π + x) = sin2x.
6) sin(π - a)sin(π/2 - a) · [ tan(π + a) - cot(2π - a) ] = 1.
7) sin(π/2 + x)cos(π/2 - x) + sin(π -x)cos(x - 2π) + tan(-x)tan(π/2 + x) = 1.
8) Knowing that tan(x) = 5/12 and that x ends in the 3rd Quadrant, calculate the numerical value of A:
A = sin(2π -x) + cos(π/2 - x) + tan(x + 2π) + cot(-x).
In Problems 10 through 15, use x = π/6, y = π/4, and z = π/3 to
calculate the numerical value of each:
In any circle of radius R, each side of a regular inscribed octagon measures as C8 given on the right. Use this length to calculate the trig ratios of 22.5° angle. Note that a regular octagon has 8 equal sides. An inscribed polygon (like the one on the right) has its corners on the circle while a circumscribed polygon has a circle inside it tangent to each of its sides (not shown here).
17) Similar to Problem 16 above, draw a regular inscribed decagon (10 equal sides) in a circle of radius R. each side of a regular inscribed decagon measures
as C10 given below. Use this length to calculate the trig ratios of 18° angle:
18) If C12 is given as shown below, calculate the trig ratios of 15° angle.