The Unit Circle: Sine and Cosine for Every Angle

The triangle definition runs out of angles

Sine and cosine were defined as side ratios in a right triangle. That definition works, and it is where the names come from. It also has a hard ceiling.

A right triangle already spends $90°$ on its right angle, so each of the other two angles has to be smaller than $90°$. Ask for $\sin 150°$ and the definition has nothing to offer: no right triangle contains a $150°$ angle.

Yet a wheel turns past $150°$ without difficulty, and its rider keeps having a height. The mathematics has to cover the turn, so the definition has to grow.

The unit circle definition

The unit circle is the circle of radius $1$ centered at the origin.

Put an angle in standard position: vertex at the origin, one side lying along the positive $x$-axis. Turn counterclockwise through the angle. The other side — the terminal side — crosses the unit circle at exactly one point.

That point is the definition.

$$(\cos\theta,\ \sin\theta) = \text{the point where the terminal side meets the unit circle}$$

Nothing stops the turn at $90°$. Keep going into the second quadrant, the third, the fourth, past a full lap, or backwards. Every real number names a turn, so every real number has a sine and a cosine.

Why the coordinates are the old ratios

The new definition does not overturn the old one. For an acute angle it repeats it.

Take an acute $\theta$ and drop a vertical from the point to the $x$-axis, as the dashed line above does. That makes a right triangle with the angle at the origin. Its horizontal leg is $x$, its vertical leg is $y$, and its hypotenuse is a radius, so the hypotenuse is $1$.

$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1} = x \qquad \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{1} = y$$

The radius of $1$ is doing all the work: it turns a ratio into a plain coordinate. The circle definition agrees with the triangle definition wherever both apply, and keeps going where the triangle cannot. That is what makes it a replacement rather than a rival.

Signs by quadrant

Sine and cosine are coordinates, so their signs are the signs of the coordinates.

Quadrant	Angles	$x = \cos\theta$	$y = \sin\theta$
I	$0$ to $\tfrac{\pi}{2}$	$+$	$+$
II	$\tfrac{\pi}{2}$ to $\pi$	$-$	$+$
III	$\pi$ to $\tfrac{3\pi}{2}$	$-$	$-$
IV	$\tfrac{3\pi}{2}$ to $2\pi$	$+$	$-$

There is no rule to memorize here. Sketch the point and read off which side of each axis it is on.

The special angles

Three angles have exact coordinates worth knowing, and all three come from the special right triangles.

For $\theta = \tfrac{\pi}{3}$ (that is $60°$), the dashed vertical makes a $30$-$60$-$90$ triangle with hypotenuse $1$. In such a triangle the short leg is half the hypotenuse and the long leg is $\tfrac{\sqrt{3}}{2}$ times it.

For $\theta = \tfrac{\pi}{4}$ the triangle is $45$-$45$-$90$, so the two legs match, and $x^2 + y^2 = 1$ forces each to be $\tfrac{\sqrt{2}}{2}$.

$\theta$	Degrees	$\cos\theta$	$\sin\theta$
$0$	$0°$	$1$	$0$
$\tfrac{\pi}{6}$	$30°$	$\tfrac{\sqrt{3}}{2}$	$\tfrac{1}{2}$
$\tfrac{\pi}{4}$	$45°$	$\tfrac{\sqrt{2}}{2}$	$\tfrac{\sqrt{2}}{2}$
$\tfrac{\pi}{3}$	$60°$	$\tfrac{1}{2}$	$\tfrac{\sqrt{3}}{2}$
$\tfrac{\pi}{2}$	$90°$	$0$	$1$

Read the two middle columns downward. Cosine falls from $1$ to $0$ and sine climbs from $0$ to $1$, because the point is sliding counterclockwise from $(1, 0)$ up to $(0, 1)$.

Reflections: π − x, π + x and 2π − x

Every other angle reduces to one of those five. The circle is symmetric, so three reflections cover the whole plane.

Reflect across the $y$-axis. The angle $\pi - x$ lands at the mirror image of the point for $x$. Mirroring across the $y$-axis negates the first coordinate and leaves the second alone.

$$\cos(\pi - x) = -\cos x \qquad \sin(\pi - x) = \sin x$$

Reflect through the origin. The angle $\pi + x$ points the opposite way, so both coordinates flip sign.

$$\cos(\pi + x) = -\cos x \qquad \sin(\pi + x) = -\sin x$$

Reflect across the $x$-axis. The angle $2\pi - x$ is the same turn measured the other way, so the height flips and the width stays.

$$\cos(2\pi - x) = \cos x \qquad \sin(2\pi - x) = -\sin x$$

The acute angle $x$ used in these rules is called the reference angle. Find it, look up its exact values, then fix the two signs from the quadrant.

Periodicity and symmetry

Add a full turn and you land where you started.

$$\cos(\theta + 2\pi) = \cos\theta \qquad \sin(\theta + 2\pi) = \sin\theta$$

Both functions are periodic with period $2\pi$. Subtract full turns until the angle sits between $0$ and $2\pi$, then work from there.

Turning backwards mirrors the point across the $x$-axis, which leaves the first coordinate alone and negates the second.

$$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta$$

In function language, cosine is even and sine is odd. Both facts are the same picture: the points for $\theta$ and $-\theta$ sit one above the other.

Tangent on the circle

Tangent was opposite over adjacent, which is $y$ over $x$.

$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

That is the slope of the terminal side. It is undefined exactly where $\cos\theta = 0$ — at $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$ — because a vertical line has no slope.

$\theta$	$\tan\theta$
$0$	$0$
$\tfrac{\pi}{6}$	$\tfrac{\sqrt{3}}{3}$
$\tfrac{\pi}{4}$	$1$
$\tfrac{\pi}{3}$	$\sqrt{3}$
$\tfrac{\pi}{2}$	undefined

Worked examples

Common mistakes

Practice problems