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Special Triangles &
Unit Circle

Master the exact trig values for \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\) using special right triangles, then extend them to every quadrant with the unit circle.

Precalculus

Why Are Certain Triangles Special?

In trigonometry, two right triangles appear so often that their side ratios are worth memorizing exactly: the 45-45-90 triangle and the 30-60-90 triangle. Their ratios give us exact values for sine, cosine, and tangent at key angles: no calculator needed.

The 45-45-90 Triangle

\(45^\circ\) \(45^\circ\) \(1\) \(1\) \(\sqrt{2}\) C B A

The 30-60-90 Triangle

\(30^\circ\) \(60^\circ\) \(\sqrt{3}\) \(1\) \(2\) C B A

Connecting to the Unit Circle

A unit circle has radius 1, centered at the origin. Any point on it has coordinates \((\cos\theta, \sin\theta)\). When we embed our special triangles inside the unit circle (with hypotenuse = 1), the legs become exact trig values.

\(\frac{\pi}{6}\) \(\cos\theta\) \(\sin\theta\) 1 1 −1 −1

Exact Values: \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\)

Angle \(\sin\theta\) \(\cos\theta\) \(\tan\theta\)
\(\frac{\pi}{6}\) (30°) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\)
\(\frac{\pi}{4}\) (45°) \(\frac{\sqrt{2}}{2}\) \(\frac{\sqrt{2}}{2}\) \(1\)
\(\frac{\pi}{3}\) (60°) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\)

Reference Angles & Quadrant Signs

A reference angle is the acute angle formed between the terminal side and the \(x\)-axis. The trig values at any angle equal the values at its reference angle, with a sign determined by the quadrant.

QI sin+, cos+ QII sin+, cos− QIII sin−, cos− QIV sin−, cos+

Symmetry Identities on the Unit Circle

\(\pi - x\) (QII)
\(\sin(\pi - x) = \sin x\)
\(\cos(\pi - x) = -\cos x\)
\(\tan(\pi - x) = -\tan x\)
\(\pi + x\) (QIII)
\(\sin(\pi + x) = -\sin x\)
\(\cos(\pi + x) = -\cos x\)
\(\tan(\pi + x) = \tan x\)
\(2\pi - x\) (QIV)
\(\sin(2\pi - x) = -\sin x\)
\(\cos(2\pi - x) = \cos x\)
\(\tan(2\pi - x) = -\tan x\)

Reference Angle vs. Symmetry Identity

Feature Reference Angle Symmetry Identity
What it gives The acute angle to the \(x\)-axis An algebraic formula relating \(f(\theta)\) to \(f(x)\)
Used for Finding specific values (e.g., \(\sin 150°\)) Proving general relationships
Output A number: the angle itself An equation with ±signs
Example ref angle of \(\frac{5\pi}{6}\) is \(\frac{\pi}{6}\) \(\sin(\pi - x) = \sin x\)

How to Find Exact Trig Values

1
Identify the quadrant of the given angle. This determines which trig functions are positive or negative.
2
Find the reference angle: the acute angle between the terminal side and the \(x\)-axis.
3
Look up the special triangle value at the reference angle (\(\frac{\pi}{6}\), \(\frac{\pi}{4}\), or \(\frac{\pi}{3}\)).
4
Apply the correct sign from the quadrant. Done: you have the exact value.

Example 1: \(\sin\!\left(\frac{5\pi}{6}\right)\)

\(\frac{\pi}{6}\) \(\frac{5\pi}{6}\) \(\frac{1}{2}\) QII
1
Quadrant: \(\frac{5\pi}{6}\) is between \(\frac{\pi}{2}\) and \(\pi\), so it is in QII where sine is positive.
2
Reference angle: \(\pi - \frac{5\pi}{6} = \frac{\pi}{6}\).
3
Special value: \(\sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2}\).
4
Apply sign: positive in QII, so \(\sin\!\left(\frac{5\pi}{6}\right) = \frac{1}{2}\). ✓

Example 2: \(\cos\!\left(\frac{5\pi}{4}\right)\)

\(\frac{\pi}{4}\) \(\frac{5\pi}{4}\) \(\cos\) QIII
1
Quadrant: \(\frac{5\pi}{4}\) is between \(\pi\) and \(\frac{3\pi}{2}\), so it is in QIII. Cosine is negative here.
2
Reference angle: \(\frac{5\pi}{4} - \pi = \frac{\pi}{4}\).
3
Special value: \(\cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\).
4
Apply sign: negative in QIII, so \(\cos\!\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}\). ✓

Example 3: \(\tan\!\left(\frac{11\pi}{6}\right)\)

\(\frac{\pi}{6}\) \(\frac{11\pi}{6}\) QIV
1
Quadrant: \(\frac{11\pi}{6}\) is between \(\frac{3\pi}{2}\) and \(2\pi\), so it is in QIV. Tangent is negative there (sin−, cos+).
2
Reference angle: \(2\pi - \frac{11\pi}{6} = \frac{\pi}{6}\).
3
Special value: \(\tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).
4
Apply sign: negative in QIV, so \(\tan\!\left(\frac{11\pi}{6}\right) = -\frac{\sqrt{3}}{3}\). ✓

Using the \(\pi - x\) Identity

Problem: Express \(\cos\!\left(\frac{5\pi}{6}\right)\) in terms of a first-quadrant value using the symmetry identity \(\cos(\pi - x) = -\cos x\).

1
Write the angle as \(\pi - x\): \(\frac{5\pi}{6} = \pi - \frac{\pi}{6}\), so \(x = \frac{\pi}{6}\).
2
Apply the identity: \(\cos(\pi - \frac{\pi}{6}) = -\cos\frac{\pi}{6}\).
3
Substitute the exact value: \(-\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}\).

Common Mistakes to Avoid

✕ Swapping sin and cos
Remember: \(\sin 30^\circ = \frac{1}{2}\) (the smaller value). Mixing up \(\sin 30^\circ\) and \(\cos 30^\circ\) is the most common error.
✕ Wrong reference angle
In QIII, the reference angle is \(\theta - \pi\), not \(\pi - \theta\). For \(\frac{7\pi}{6}\): ref = \(\frac{7\pi}{6} - \pi = \frac{\pi}{6}\).
✕ Forgetting the sign
Getting the magnitude right but leaving off the negative sign. Always check the quadrant before writing your final answer.

Summary: Symmetry Identities

Identity Form \(\sin\) \(\cos\) \(\tan\)
\(\pi - x\) \(+\sin x\) \(-\cos x\) \(-\tan x\)
\(\pi + x\) \(-\sin x\) \(-\cos x\) \(+\tan x\)
\(2\pi - x\) \(-\sin x\) \(+\cos x\) \(-\tan x\)

Question 1 of 5

What is the exact value of \(\sin\!\left(\frac{\pi}{3}\right)\)?

Question 2 of 5

What is \(\cos\!\left(\frac{3\pi}{4}\right)\)?

Question 3 of 5

Using the identity \(\sin(\pi + x) = -\sin x\), what is \(\sin\!\left(\frac{7\pi}{6}\right)\)?

Question 4 of 5

What is \(\tan\!\left(\frac{5\pi}{3}\right)\)?

Question 5 of 5

In the 45-45-90 triangle, the ratio of a leg to the hypotenuse is: