Exact Values Of Trigonometric Functions
Master the exact values of sine, cosine and tangent for Queensland Year 11 Mathematical Methods (QCAA). At the special angles these ratios are surds read straight from the two special triangles.
You will learn to recall these values, keep answers exact with surds rather than rounding, and extend them beyond the first quadrant using reference angles and the ASTC sign rule — a fast, calculator-free skill relied on across the QCAA course.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the exact values of sine, cosine and tangent at \(0,\dfrac{\pi}{6},\dfrac{\pi}{4},\dfrac{\pi}{3},\dfrac{\pi}{2}\) (that is \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\)) come from the two special triangles and the unit circle. This page shows how to read and use them, keeping surds exact, and how to extend them to other angles with a reference angle and ASTC.
The exact values are the surd forms of the trig ratios at the special angles, not rounded decimals. They come from two triangles: the \(45\text{-}45\text{-}90\) triangle (legs \(1,1\), hypotenuse \(\sqrt2\)) and the \(30\text{-}60\text{-}90\) triangle (sides \(1,\sqrt3,2\)).
Reading SOH CAH TOA off these triangles gives \(\sin 30^\circ=\dfrac{1}{2}\), \(\cos 30^\circ=\dfrac{\sqrt3}{2}\), \(\tan 45^\circ=1\), \(\tan 60^\circ=\sqrt3\), and so on. The same values sit on the unit circle as the coordinates of the special-angle points.
For an angle outside the first quadrant, use the reference angle to get the size and ASTC for the sign. The Pythagorean identity \(\sin^2 t+\cos^2 t=1\) recovers one exact ratio from another when the quadrant is known.
The first-quadrant exact values:
| \(\theta\) | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) |
|---|---|---|---|
| \(0\) | \(0\) | \(1\) | \(0\) |
| \(\dfrac{\pi}{6}\ (30^\circ)\) | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt3}{2}\) | \(\dfrac{\sqrt3}{3}\) |
| \(\dfrac{\pi}{4}\ (45^\circ)\) | \(\dfrac{\sqrt2}{2}\) | \(\dfrac{\sqrt2}{2}\) | \(1\) |
| \(\dfrac{\pi}{3}\ (60^\circ)\) | \(\dfrac{\sqrt3}{2}\) | \(\dfrac{1}{2}\) | \(\sqrt3\) |
| \(\dfrac{\pi}{2}\ (90^\circ)\) | \(1\) | \(0\) | undefined |
The Pythagorean identity, used to recover one ratio from another:
How to write an exact value
- Identify the special angle (or its reference angle) and recall the triangle or table entry.
- Pick the ratio with SOH CAH TOA, or read the coordinate from the unit circle.
- Attach the sign from the quadrant (ASTC) and leave surds exact.
For the \(60^\circ\) angle: opposite \(=\sqrt3\), adjacent \(=1\), hypotenuse \(=2\). Sine:
| \(\sin 60^\circ\) | \(=\) | \(\dfrac{\text{opp}}{\text{hyp}}\) |
| \(=\) | \(\dfrac{\sqrt3}{2}\) |
Cosine:
| \(\cos 60^\circ\) | \(=\) | \(\dfrac{\text{adj}}{\text{hyp}}\) |
| \(=\) | \(\dfrac{1}{2}\) |
Tangent:
| \(\tan 60^\circ\) | \(=\) | \(\dfrac{\text{opp}}{\text{adj}}\) |
| \(=\) | \(\sqrt3\) |
\(\sin 60^\circ=\dfrac{\sqrt3}{2},\ \cos 60^\circ=\dfrac{1}{2},\ \tan 60^\circ=\sqrt3\).
The legs are \(1,1\) and the hypotenuse is \(\sqrt2\). Sine (rationalising the surd):
| \(\sin\dfrac{\pi}{4}\) | \(=\) | \(\dfrac{1}{\sqrt2}\) |
| \(=\) | \(\dfrac{\sqrt2}{2}\) |
By symmetry cosine equals sine here:
| \(\cos\dfrac{\pi}{4}\) | \(=\) | \(\dfrac{\sqrt2}{2}\) |
Tangent is opposite over adjacent:
| \(\tan\dfrac{\pi}{4}\) | \(=\) | \(\dfrac{1}{1}\) |
| \(=\) | \(1\) |
\(\sin\dfrac{\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\) and \(\tan\dfrac{\pi}{4}=1\).
\(\dfrac{5\pi}{6}\) is in the second quadrant, where cosine is negative. Reference angle:
| \(\text{ref}\) | \(=\) | \(\pi-\dfrac{5\pi}{6}\) |
| \(=\) | \(\dfrac{\pi}{6}\) |
Attach the second-quadrant sign to the table value \(\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\):
| \(\cos\dfrac{5\pi}{6}\) | \(=\) | \(-\cos\dfrac{\pi}{6}\) |
| \(=\) | \(-\dfrac{\sqrt3}{2}\) |
\(\cos\dfrac{5\pi}{6}=-\dfrac{\sqrt3}{2}\).
Use \(\sin^2 t+\cos^2 t=1\), so \(\sin^2 t=1-\cos^2 t\):
| \(\sin^2 t\) | \(=\) | \(1-\left(\dfrac{3}{5}\right)^2\) |
| \(=\) | \(1-\dfrac{9}{25}\) | |
| \(=\) | \(\dfrac{16}{25}\) |
Take the square root:
| \(\sin t\) | \(=\) | \(\pm\dfrac{4}{5}\) |
In the fourth quadrant sine is negative, so choose the negative sign.
\(\sin t=-\dfrac{4}{5}\).
Common pitfalls
Frequently asked questions
What is the exact value of sin 30 degrees?
\(\sin 30^\circ=\dfrac{1}{2}\), read from the \(30\text{-}60\text{-}90\) triangle as opposite over hypotenuse (\(1\) over \(2\)).
Where do the exact values come from?
From two special triangles: the \(45\text{-}45\text{-}90\) (\(1,1,\sqrt2\)) and the \(30\text{-}60\text{-}90\) (\(1,\sqrt3,2\)). SOH CAH TOA on these gives the surd values.
How do you find an exact value outside the first quadrant?
Use the reference angle for the size and ASTC for the sign. For example \(\cos\dfrac{5\pi}{6}=-\cos\dfrac{\pi}{6}=-\dfrac{\sqrt3}{2}\).
What is tan 90 degrees?
It is undefined, because \(\cos 90^\circ=0\) and \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) then has a zero denominator.
How does sin squared plus cos squared equal 1 help?
If you know one ratio and the quadrant, it gives the other exactly: \(\sin^2 t=1-\cos^2 t\), then choose the sign from the quadrant.