Trigonometric Ratios of Special Angles

Trigonometric ratios of 45°

To find the trigonometric ratios of a 45° angle, we can consider the following isosceles right triangle:

This triangle has two angles with the same measures, so both of its legs have the same length of 1 unit.

We can use the Pythagorean theorem to find the length of the hypotenuse:

$latex {{AB}^2}={{BC}^2}+{{AC}^2}$

$latex {{AB}^2}={{1}^2}+{{1}^2}$

$latex {{AB}^2}=2$

$latex AB=\sqrt{2}$

Now, we can use the same triangle to obtain the values of the trigonometric ratios, sine, cosine, and tangent:

Using the definition of sine (opposite side over hypotenuse), we have:

$$\sin(45^{\circ})=\frac{AC}{AB}=\frac{1}{\sqrt{2}}$$

Using the definition of cosine (adjacent side over hypotenuse), we have:

$$\cos(45^{\circ})=\frac{BC}{AB}=\frac{1}{\sqrt{2}}$$

Using the definition of the tangent (opposite side over adjacent side), we have:

$$\tan(45^{\circ})=\frac{AC}{BC}=1$$

In short, we have the following:

$$\sin(45^{\circ})=\frac{1}{\sqrt{2}}$$ $$\cos(45^{\circ})=\frac{1}{\sqrt{2}}$$ $$\tan(45^{\circ})=1$$

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Trigonometric ratios of 30°

To find the trigonometric ratios of the 30° angle, we are going to use an equilateral triangle that has side lengths of 2 units.

D is the point where the perpendicular segment from A meets the base. The segment AD divides the equilateral triangle into two equal triangles that have the angles 30°, 60° and 90°.

We can find the length of segment AD using the Pythagorean theorem:

$latex {{AD}^2}+{{BD}^2}={{AB}^2}$

$latex {{AD}^2}+{{1}^2}={{2}^2}$

$latex {{AD}^2}+1=4$

$latex {{AD}^2}=3$

$latex AD=\sqrt{3}$

Using the definition of sine (opposite side over hypotenuse), we have:

$$\sin(30^{\circ})=\frac{BD}{AB}=\frac{1}{2}$$

Using the definition of cosine (adjacent side over hypotenuse), we have:

$$\cos(30^{\circ})=\frac{AD}{AB}=\frac{\sqrt{3}}{2}$$

Using the definition of the tangent (opposite side over adjacent side), we have:

$$\tan(30^{\circ})=\frac{BD}{AD}=\frac{1}{\sqrt{3}}$$

In short, we have the following:

$$\sin(30^{\circ})=\frac{1}{2}$$ $$\cos(30^{\circ})=\frac{\sqrt{3}}{2}$$ $$\tan(30^{\circ})=\frac{1}{\sqrt{3}}$$

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Trigonometric ratios of 60°

We can find the trigonometric ratios of the 60° angle using the same triangle that we used to find the trigonometric ratios of 30°.

Therefore, we use the lengths $latex AB=2$, $latex BD=1$, and $latex AD=\sqrt{3}$ with the trigonometric functions sine, cosine, and tangent.

Using the definition of sine (opposite side over hypotenuse), we have:

$$\sin(60^{\circ})=\frac{AD}{AB}=\frac{\sqrt{3}}{2}$$

Using the definition of cosine (adjacent side over hypotenuse), we have:

$$\cos(60^{\circ})=\frac{BD}{AB}=\frac{1}{2}$$

Using the definition of the tangent (opposite side over adjacent side), we have:

$$\tan(60^{\circ})=\frac{AD}{BD}=\sqrt{3}$$

In short, we have the following:

$$\sin(60^{\circ})=\frac{\sqrt{3}}{2}$$ $$\cos(60^{\circ})=\frac{1}{2}$$ $$\tan(60^{\circ})=\sqrt{3}$$

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Table of the trigonometric ratios of special angles

The trigonometric ratios of the angles 30°, 45° and 60° are summarized in the following table:

Ratio	θ = 30°	θ = 30°	θ = 30°
sin θ	$latex \frac{1}{2}$	$latex \frac{1}{\sqrt{2}}$	$latex \frac{\sqrt{3}}{2}$
cos θ	$latex \frac{\sqrt{3}}{2}$	$latex \frac{1}{\sqrt{2}}$	$latex \frac{1}{2}$
tan θ	$latex \frac{1}{\sqrt{3}}$	1	$latex \sqrt{3}$

It is a good idea to memorize these trigonometric ratios since several problems found in trigonometry use the 30°, 45°, and 60° angles.

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See also

Interested in learning more about special angles and triangles? Look at these pages:

• Special Triangles – Formulas and Examples
• Angle Between Two Lines – Formula and Examples
• Triangle Inequality – Definition, Proof and Examples
• Similar and Congruent Triangles – Criteria