Rule

Trigonometric Ratios of Notable Angles

The trigonometric ratios of some notable angles are usually known by heart.
θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ 0 1/2 sqrt(2)/2 sqrt(3)/2 1
cos θ 1 sqrt(3)/2 sqrt(2)/2 1/2 0
tan θ 0 sqrt(3)/3 1 sqrt(3) undefined
The above values can be calculated by either using a calculator or by using the unit circle and the Pythagorean Theorem.

Proof

This proof will be divided into three parts. Each of them will correspond to a different group of angles and their trigonometric ratios.

  1. 90- and 0-degree angles
  2. 45-degree angles
  3. 30- and 60-degree angles

The first part will use the unit circle, while the second and third parts will use triangles.

90- and 0-Degree Angles

In this section, the value of sine, cosine, and tangent of θ= 90^(∘) will be calculated. The cosine and sine of an angle in standard position are the first and second coordinates, respectively, of the point of intersection of the terminal side of the angle and the unit circle.

circleandangle

The terminal side of the angle and the unit circle intersect at (0,1). Therefore, the cosine and sine of 90^(∘) are 0 and 1, respectively. Since the tangent of an angle equals sine over cosine, the value of the tangent of 90^(∘) is 1 ÷ 0 and is therefore undefined, as division by zero is not possible. sin 90^(∘)&=1 cos 90^(∘)&=0 tan 90^(∘)&=undefined Next, the trigonometric ratios of 0^(∘) will be calculated. The terminal side of this angle lies on the x-axis.

circleandangle

The terminal side of the angle and the unit circle intersect at (1,0). Therefore, the cosine and sine of 0^(∘) are 1 and 0, respectively. Since the tangent of an angle equals sine over cosine, the value of the tangent of 0^(∘) is 0÷ 1=0. sin 0^(∘)&=0 cos 0^(∘)&=1 tan 0^(∘)&=0

45-Degree Angle

To find the trigonometric ratios of θ=45^(∘), a right isosceles triangle with hypotenuse 1 will be drawn. Since the right angle measures 90^(∘), by the Triangle Angle Sum Theorem, the acute angles measure 45^(∘). Let x be the length of the legs of the right triangle.

isosceles triangle
By the Pythagorean Theorem, the sum of the squares of lengths of the legs is equal to the length of the hypotenuse squared.
x^2+x^2=1^2
Solve for x
2x^2=1^2
2x^2=1
x^2=1/2
x=sqrt(1/2)
x=sqrt(1)/sqrt(2)
x=1/sqrt(2)
x=1sqrt(2)/sqrt(2)sqrt(2)
x=sqrt(2)/sqrt(2)sqrt(2)
x=sqrt(2)/2
The length of the legs of the triangle is sqrt(2)2.
isosceles triangle

The sine of the acute angle of a right triangle is defined as the quotient of the lengths of the side opposite to the angle and the hypotenuse of the right triangle. Similarly, the cosine is defined as the quotient of the lengths of the side adjacent to the angle and the hypotenuse of the right triangle. cc sin 45^(∘)=sqrt(2)2/1 & cos 45^(∘)=sqrt(2)2/1 ⇓ & ⇓ sin 45^(∘)=sqrt(2)/2 & cos 45^(∘)=sqrt(2)/2 Finally, the tangent of an angle is the quotient of the sine and the cosine of the angle. Since any number divided by itself is 1, the tangent of 45^(∘) is 1. sin 45^(∘)&=sqrt(2)/2 [0.8em] cos 45^(∘)&=sqrt(2)/2 [0.8em] tan 45^(∘)&=1

30- and 60-Degree Angles

Consider an equilateral triangle with a side length of 1.

equilateral triangle

The altitude of this type of a triangle bisects the base and its opposite angle. Consider one of the right triangles obtained. This triangle has a hypotenuse length of 1, base length of 12, and angles with measures 30^(∘), 60^(∘), and 90^(∘).

equilateral triangle
The value of the altitude h can be found by using the Pythagorean Theorem.
( 1/2)^2+ h^2= 1^2
Solve for h
1^2/2^2+h^2=1^2
1/4+h^2=1
h^2=1-1/4
h^2=4/4-1/4
h^2=3/4
h=sqrt(3/4)
h=sqrt(3)/sqrt(4)
h=sqrt(3)/2
The obtained value of h can be added to the right triangle.
equilateral triangle

Finally, the sine, cosine, and tangent of 30^(∘) and 60^(∘) can be obtained by using the definitions of the trigonometric ratios.

Definition Substitute Simplify
sin 30^(∘) opp/hyp 12/1 1/2
cos 30^(∘) adj/hyp sqrt(3)2/1 sqrt(3)/2
tan 30^(∘) opp/adj 12/sqrt(3)2 sqrt(3)/3
sin 60^(∘) opp/hyp sqrt(3)2/1 sqrt(3)/2
cos 60^(∘) adj/hyp 12/1 1/2
tan 60^(∘) opp/adj sqrt(3)2/12 sqrt(3)

With the obtained results, the sine, cosine, and tangent of all five notable angles were obtained.

θ=0^(∘) θ=30^(∘) θ=45^(∘) θ=60^(∘) θ=90^(∘)
sin θ 0 1/2 sqrt(2)/2 sqrt(3)/2 1
cos θ 1 sqrt(3)/2 sqrt(2)/2 1/2 0
tan θ 0 sqrt(3)/3 1 sqrt(3) undefined

Extra

Constructing the Table
The following five steps help to construct the table for notable angles.
1
Write the Notable Angles and Trigonometric Functions
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First, the heading row with the notable angles will be written. The first column containing sin θ and cos θ will also be written.

θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ
cos θ
2
Filling the Sine and Cosine Rows
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In the sine row, the integer numbers from 0 to 4 will be written one per column. In the cosine row, the same numbers but in the opposite order will be written.

θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ 0 1 2 3 4
cos θ 4 3 2 1 0
3
Take the Square Root of Each Number
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Now, the square root of each number will be calculated.

θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ sqrt(0)=0 sqrt(1)=1 sqrt(2) sqrt(3) sqrt(4)=2
cos θ sqrt(4)=2 sqrt(3) sqrt(2) sqrt(1)=1 sqrt(0)=0
4
Divide Each Number by 2
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Each number will now be divided by 2.

θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ 0/2=0 1/2 sqrt(2)/2 sqrt(3)/2 2/2=1
cos θ 2/2=1 sqrt(3)/2 sqrt(2)/2 1/2 0/2=0
5
Writing the Tangent Row
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Finally, to write the third row, corresponding to the tangent ratio, the fact that tan θ = sinθcosθ will be used. The number in the sine row will be divided by the number in the cosine row.

θ=0^(∘)
or
θ=0rad
θ=30^(∘)
or
θ=π/6rad
θ=45^(∘)
or
θ=π/4rad
θ=60^(∘)
or
θ=π/3rad
θ=90^(∘)
or
θ=π/2rad
sin θ 0 1/2 sqrt(2)/2 sqrt(3)/2 1
cos θ 1 sqrt(3)/2 sqrt(2)/2 1/2 0
tan θ 0/1=0 12/sqrt(3)2=sqrt(3)/3 sqrt(2)2/sqrt(2)2=1 sqrt(3)2/12=sqrt(3) 1/0⇒ undefined
Exercises