Most angles have messy sines and cosines. A few have exact values you can find with no calculator, because they come from the two special right triangles. Put each triangle inside the unit circle, with its hypotenuse as a radius of length $1$.
Why it works
$45^\circ$: the two legs are equal, say $x$. The Pythagorean theorem gives $x^2 + x^2 = 1$, so $x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
$30^\circ$ and $60^\circ$: a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $1$ is half of an equilateral triangle with sides $1$, so its short side is $\frac{1}{2}$. The other leg is $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. The short side is always across from the small $30^\circ$ angle.
$30^\circ$ and $60^\circ$: a $30^\circ$-$60^\circ$-$90^\circ$ triangle with hypotenuse $1$ is half of an equilateral triangle with sides $1$, so its short side is $\frac{1}{2}$. The other leg is $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. The short side is always across from the small $30^\circ$ angle.
| $\theta$ | $0$ | $\frac{\pi}{6} = 30^\circ$ | $\frac{\pi}{4} = 45^\circ$ | $\frac{\pi}{3} = 60^\circ$ | $\frac{\pi}{2} = 90^\circ$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\frac{\sqrt{3}}{3}$ | $1$ | $\sqrt{3}$ | undefined |
Key idea
At $30^\circ$ the point on the unit circle is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$; at $60^\circ$ it is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$; at $45^\circ$ it is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$.
Worked examples
Worked example: A sine
Find $\sin 60^\circ$.
- At $60^\circ$ the radius is steep, so the long leg, $\frac{\sqrt{3}}{2}$, is the vertical one.
- Sine is the height: $\sin 60^\circ = \frac{\sqrt{3}}{2}$.
Answer: $\frac{\sqrt{3}}{2} \approx 0.866$
Worked example: A tangent
Find $\tan\frac{\pi}{6}$.
- $\frac{\pi}{6} = 30^\circ$, where $\sin = \frac{1}{2}$ and $\cos = \frac{\sqrt{3}}{2}$.
- $\tan\frac{\pi}{6} = \dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}$.
- Rationalize: $\dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.
Answer: $\frac{\sqrt{3}}{3}$
Worked example: Putting values together
Find the exact value of $2\sin 30^\circ \cos 30^\circ$.
- $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
- $2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.
Answer: $\frac{\sqrt{3}}{2}$ (which is also $\sin 60^\circ$, a hint of things to come)
Common mistake
Swapping $\sin 30^\circ$ and $\sin 60^\circ$, writing $\sin 30^\circ = \frac{\sqrt{3}}{2}$. Check with the picture: as the angle grows from $0^\circ$ to $90^\circ$, the point climbs, so sine grows. $\sin 30^\circ$ must be the smaller value, $\frac{1}{2}$, and $\frac{\sqrt{3}}{2} \approx 0.87$ belongs to $60^\circ$.
A memory pattern: $$\sin 0^\circ, \sin 30^\circ, \sin 45^\circ, \sin 60^\circ, \sin 90^\circ$$ are $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$. Cosine runs the same list backward.