Trigonometric Ratios of Notable Angles

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

$90 -$ and $0 -$ degree angles
$45 -$ degree angles
$30 -$ 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 $θ = 9 0^{\circ}$ 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.

The terminal side of the angle and the unit circle intersect at

(0, 1) .

Therefore, the cosine and sine of

9 0^{\circ}

are

0

and

1,

respectively. Since the tangent of an angle equals sine over cosine, the value of the tangent of

9 0^{\circ}

1 \div 0

and is therefore undefined, as division by zero is not possible.

sin 9 0^{\circ} cos 9 0^{\circ} tan 9 0^{\circ} = 1 = 0 = undefined

Next, the trigonometric ratios of

0^{\circ}

will be calculated. The terminal side of this angle lies on the

x -

axis.

The terminal side of the angle and the unit circle intersect at

(1, 0) .

Therefore, the cosine and sine of

0^{\circ}

are

1

and

0,

respectively. Since the tangent of an angle equals sine over cosine, the value of the tangent of

0^{\circ}

0 \div 1 = 0 .

sin 0^{\circ} cos 0^{\circ} tan 0^{\circ} = 0 = 1 = 0

$45 -$ Degree Angle

To find the trigonometric ratios of $θ = 4 5^{\circ},$ a right isosceles triangle with hypotenuse $1$ will be drawn. Since the right angle measures $9 0^{\circ},$ by the Triangle Angle Sum Theorem, the acute angles measure $4 5^{\circ} .$ Let $x$ be the length of the legs of the right 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

AddTerms

Add terms

2 x^{2} = 1^{2}

BaseOne

$1^{a} = 1$

2 x^{2} = 1

DivEqn

$LHS / 2 = RHS / 2$

x^{2} = \frac{1}{2}

SqrtEqn

$LHS = RHS$

x = \frac{1}{2}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

x = \frac{1}{2}

CalcRoot

Calculate root

x = \frac{1}{2}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

x = \frac{1 2}{2 2}

IdPropMult

\IdPropMult

x = \frac{2}{2 2}

MultSqrt

$a \cdot a = a$

x = \frac{2}{2}

The length of the legs of the triangle is

\frac{2}{2} .

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.

sin 4 5^{\circ} = \frac{\frac{2}{2}}{1} ⇓ sin 4 5^{\circ} = \frac{2}{2} cos 4 5^{\circ} = \frac{\frac{2}{2}}{1} ⇓ cos 4 5^{\circ} = \frac{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

4 5^{\circ}

1 .

sin 4 5^{\circ} cos 4 5^{\circ} tan 4 5^{\circ} = \frac{2}{2} = \frac{2}{2} = 1

$30 -$ and $60 -$ Degree Angles

Consider an equilateral triangle with a side length of $1 .$

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 $\frac{1}{2},$ and angles with measures $3 0^{\circ},$ $6 0^{\circ},$ and $9 0^{\circ} .$

The value of the altitude

h

can be found by using the Pythagorean Theorem.

(\frac{1}{2})^{2} + h^{2} = 1^{2}

Solve for

h

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{1 ^{2}}{2 ^{2}} + h^{2} = 1^{2}

CalcPow

Calculate power

\frac{1}{4} + h^{2} = 1

SubEqn

$LHS - \frac{1}{4} = RHS - \frac{1}{4}$

h^{2} = 1 - \frac{1}{4}

OneToFrac

Rewrite $1$ as $\frac{4}{4}$

h^{2} = \frac{4}{4} - \frac{1}{4}

SubFrac

Subtract fractions

h^{2} = \frac{3}{4}

SqrtEqn

$LHS = RHS$

h = \frac{3}{4}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

h = \frac{3}{4}

CalcRoot

Calculate root

h = \frac{3}{2}

The obtained value of

h

can be added to the right triangle.

Finally, the sine, cosine, and tangent of $3 0^{\circ}$ and $6 0^{\circ}$ can be obtained by using the definitions of the trigonometric ratios.

	Definition	Substitute	Simplify
$sin 3 0^{\circ}$	$\frac{opp}{hyp}$	$\frac{\frac{1}{2}}{1}$	$\frac{1}{2}$
$cos 3 0^{\circ}$	$\frac{adj}{hyp}$	$\frac{\frac{3}{2}}{1}$	$\frac{3}{2}$
$tan 3 0^{\circ}$	$\frac{opp}{adj}$	$\frac{\frac{1}{2}}{\frac{3}{2}}$	$\frac{3}{3}$
$sin 6 0^{\circ}$	$\frac{opp}{hyp}$	$\frac{\frac{3}{2}}{1}$	$\frac{3}{2}$
$cos 6 0^{\circ}$	$\frac{adj}{hyp}$	$\frac{\frac{1}{2}}{1}$	$\frac{1}{2}$
$tan 6 0^{\circ}$	$\frac{opp}{adj}$	$\frac{\frac{3}{2}}{\frac{1}{2}}$	$3$

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

	$θ = 0^{\circ}$	$θ = 3 0^{\circ}$	$θ = 4 5^{\circ}$	$θ = 6 0^{\circ}$	$θ = 9 0^{\circ}$
$sin θ$	$0$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$	$1$
$cos θ$	$1$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$	$0$
$tan θ$	$0$	$\frac{3}{3}$	$1$	$3$	undefined

	$θ = 0^{\circ}$ or $θ = 0 rad$	$θ = 3 0^{\circ}$ or $θ = \frac{π}{6} rad$	$θ = 4 5^{\circ}$ or $θ = \frac{π}{4} rad$	$θ = 6 0^{\circ}$ or $θ = \frac{π}{3} rad$	$θ = 9 0^{\circ}$ or $θ = \frac{π}{2} rad$
$sin θ$
$cos θ$

	$θ = 0^{\circ}$ or $θ = 0 rad$	$θ = 3 0^{\circ}$ or $θ = \frac{π}{6} rad$	$θ = 4 5^{\circ}$ or $θ = \frac{π}{4} rad$	$θ = 6 0^{\circ}$ or $θ = \frac{π}{3} rad$	$θ = 9 0^{\circ}$ or $θ = \frac{π}{2} rad$
$sin θ$	$0$	$1$	$2$	$3$	$4$
$cos θ$	$4$	$3$	$2$	$1$	$0$

	$θ = 0^{\circ}$ or $θ = 0 rad$	$θ = 3 0^{\circ}$ or $θ = \frac{π}{6} rad$	$θ = 4 5^{\circ}$ or $θ = \frac{π}{4} rad$	$θ = 6 0^{\circ}$ or $θ = \frac{π}{3} rad$	$θ = 9 0^{\circ}$ or $θ = \frac{π}{2} rad$
$sin θ$	$0 = 0$	$1 = 1$	$2$	$3$	$4 = 2$
$cos θ$	$4 = 2$	$3$	$2$	$1 = 1$	$0 = 0$

	$θ = 0^{\circ}$ or $θ = 0 rad$	$θ = 3 0^{\circ}$ or $θ = \frac{π}{6} rad$	$θ = 4 5^{\circ}$ or $θ = \frac{π}{4} rad$	$θ = 6 0^{\circ}$ or $θ = \frac{π}{3} rad$	$θ = 9 0^{\circ}$ or $θ = \frac{π}{2} rad$
$sin θ$	$\frac{0}{2} = 0$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$	$\frac{2}{2} = 1$
$cos θ$	$\frac{2}{2} = 1$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$	$\frac{0}{2} = 0$

{{ article.displayTitle }}

Rule

Trigonometric Ratios of Notable Angles

Proof

$90 -$ and $0 -$ Degree Angles

$45 -$ Degree Angle

$30 -$ and $60 -$ Degree Angles

Extra

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Proof

90- and 0-Degree Angles

45-Degree Angle

30- and 60-Degree Angles

Extra

$90 -$ and $0 -$ Degree Angles

$45 -$ Degree Angle

$30 -$ and $60 -$ Degree Angles