Take a right triangle, shrink its hypotenuse to length one, and spin it around a fixed point. Suddenly every angle — even a flat one, even a backwards one — names a single point on a circle, and that point is its cosine and sine.
Angles that don't fit in a triangle
In a right triangle, an angle has to be smaller than \( 90^\circ \) — the three angles add to \( 180^\circ \), and one of them is already the right angle. That was enough to define \( \sin \), \( \cos \), and \( \tan \) from the sides of the triangle. But the world is full of angles that don't fit inside a triangle at all: a clock hand sweeping past the half-hour, a wheel rolling a full turn, a pendulum swinging back the other way.
To handle these we stop thinking of an angle as a corner of a triangle and start thinking of it as an amount of rotation. Stand a ray pointing right, then rotate it counter-clockwise. A quarter turn is \( 90^\circ \), a half turn is \( 180^\circ \), a full turn is \( 360^\circ \) — and nothing stops you from rotating further, or rotating backwards into negative angles. We need a definition of sine and cosine that survives all of that, and the circle gives us one.
The circle of radius one
Draw a circle centred at the origin with radius exactly \( 1 \). This is the unit circle. Now take your rotating ray, measure the angle \( t \) counter-clockwise from the positive \( x \)-axis, and look at where the ray crosses the circle. That single crossing point carries both of our functions:
\[ \cos t = x, \qquad \sin t = y \]In words: the cosine is the horizontal coordinate of the point, and the sine is the vertical coordinate. That is the whole definition. There is no triangle to set up, no hypotenuse to divide by — because the radius is \( 1 \), the "hypotenuse" is already \( 1 \), and \( \cos t = x/1 \) and \( \sin t = y/1 \) collapse to just \( x \) and \( y \).
For acute angles this agrees perfectly with the triangle definition you already know: drop a vertical line from the point to the \( x \)-axis and you get a right triangle with hypotenuse \( 1 \), horizontal leg \( \cos t \), and vertical leg \( \sin t \). The circle simply lets the same point keep moving past \( 90^\circ \), into the regions where \( x \) or \( y \) turn negative.
Drag the dot around the circle below. Watch the angle, the coordinates, and the two coloured projections — the horizontal one is \( \cos t \), the vertical one is \( \sin t \).
In words The point on the unit circle is always written \( (\cos t,\ \sin t) \) — cosine first, because cosine is the x-coordinate and \( x \) comes first. A handy way to remember which is which: cosine lies along the horizon.
Radians: measuring angles by arc length
Degrees are a human invention — someone long ago chopped a full turn into \( 360 \) pieces. Mathematics prefers a measure built into the circle itself: the radian. Instead of counting arbitrary slices, a radian asks how far you have travelled along the circle's edge.
On the unit circle the radius is \( 1 \), so the whole circumference is \( 2\pi \). One radian is the angle that sweeps out an arc of length \( 1 \). Sweeping all the way around covers an arc of \( 2\pi \), so a full turn is \( 2\pi \) radians. That gives the bridge between the two systems:
\[ 360^\circ = 2\pi \text{ rad}, \qquad\text{so}\qquad 180^\circ = \pi \text{ rad} \]From the half-turn equality \( 180^\circ = \pi \) you can convert either direction. To turn degrees into radians, multiply by \( \dfrac{\pi}{180} \); to turn radians into degrees, multiply by \( \dfrac{180}{\pi} \):
\[ 90^\circ = 90 \times \frac{\pi}{180} = \frac{\pi}{2}, \qquad \frac{\pi}{3} = \frac{\pi}{3} \times \frac{180}{\pi} = 60^\circ \]From here on we will quote angles in radians whenever it keeps the algebra clean, and the widget above always shows both forms side by side so you can build the dictionary in your head.
Tip You don't need a calculator to convert the common angles — just anchor on \( 180^\circ = \pi \) and scale. Half of that is \( 90^\circ = \tfrac{\pi}{2} \); a third is \( 60^\circ = \tfrac{\pi}{3} \); a sixth is \( 30^\circ = \tfrac{\pi}{6} \). Notice the smaller the angle, the bigger the number underneath \( \pi \).
The special angles and their coordinates
A handful of angles appear so often that their exact coordinates are worth memorising. They come from the \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) and \( 45^\circ\text{-}45^\circ\text{-}90^\circ \) triangles you met in right-triangle trig, now read straight off the circle as \( (\cos t,\ \sin t) \):
\[ \begin{aligned} 0 &: (1,\ 0) &\quad \tfrac{\pi}{6}=30^\circ &: \left(\tfrac{\sqrt{3}}{2},\ \tfrac{1}{2}\right) &\quad \tfrac{\pi}{4}=45^\circ &: \left(\tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{2}}{2}\right) \\[4pt] \tfrac{\pi}{3}=60^\circ &: \left(\tfrac{1}{2},\ \tfrac{\sqrt{3}}{2}\right) &\quad \tfrac{\pi}{2}=90^\circ &: (0,\ 1) & & \end{aligned} \]Read across from \( 0 \) to \( 90^\circ \) and notice the pattern: the \( x \)-coordinate slides down from \( 1 \) to \( 0 \) while the \( y \)-coordinate climbs from \( 0 \) to \( 1 \). At the exact middle, \( 45^\circ \), they are equal, because the point sits on the diagonal line \( y = x \).
The other three quadrants need no new memorising — they are mirror images. The same five points reflect across the axes, flipping a sign each time they cross. For instance, \( 120^\circ = \tfrac{2\pi}{3} \) is the mirror of \( 60^\circ \) across the \( y \)-axis, so its coordinates are \( \left(-\tfrac{1}{2},\ \tfrac{\sqrt{3}}{2}\right) \): the \( x \) flipped negative, the \( y \) stayed put. Step the widget around all four quadrants and watch the coordinates pick up their minus signs.
Signs in each quadrant
Because \( \cos t \) and \( \sin t \) are literally the coordinates of the point, their signs are decided by which quadrant the point lands in — exactly the same plus/minus pattern you learned for the coordinate plane.
- Quadrant I (\( 0 \) to \( 90^\circ \)): \( x>0 \) and \( y>0 \), so all three of \( \sin, \cos, \tan \) are positive.
- Quadrant II (\( 90^\circ \) to \( 180^\circ \)): \( x<0,\ y>0 \), so only \( \sin \) is positive.
- Quadrant III (\( 180^\circ \) to \( 270^\circ \)): \( x<0,\ y<0 \), so only \( \tan = \tfrac{\sin}{\cos} \) is positive (two negatives divide to a positive).
- Quadrant IV (\( 270^\circ \) to \( 360^\circ \)): \( x>0,\ y<0 \), so only \( \cos \) is positive.
The first letters spell a memory hook running anticlockwise from Quadrant I: All, Sine, Tangent, Cosine — often recited as "All Students Take Calculus." Whatever is named in a quadrant is the function that stays positive there; everything else is negative.
The Pythagorean identity
Here is the payoff that makes the whole construction worth it. Every point \( (x, y) \) on the unit circle satisfies the circle's own equation, \( x^2 + y^2 = 1 \). But \( x \) is just \( \cos t \) and \( y \) is just \( \sin t \), so substituting gives the single most important identity in trigonometry:
\[ \sin^2 t + \cos^2 t = 1 \]It holds for every angle \( t \), in every quadrant, with no exceptions — because every point of the circle is one unit from the centre, by definition. It is really the Pythagorean theorem applied to the little right triangle with legs \( \cos t \) and \( \sin t \) and hypotenuse \( 1 \), which is why it carries that name.
This one equation lets you find a sine from a cosine (or vice versa) whenever you know the quadrant to fix the sign. Check it on a special angle: at \( 30^\circ \), \( \sin^2 30^\circ + \cos^2 30^\circ = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{1}{4} + \tfrac{3}{4} = 1 \). It works, exactly, every time.
- Convert by multiplying by \( \tfrac{\pi}{180} \): \( 135 \times \tfrac{\pi}{180} = \tfrac{135\pi}{180} = \tfrac{3\pi}{4} \).
- The angle \( 135^\circ \) is in Quadrant II, the mirror of \( 45^\circ \) across the \( y \)-axis, so the coordinates of \( 45^\circ \) keep their sizes but the \( x \) turns negative.
- Since \( 45^\circ \) gives \( \left(\tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{2}}{2}\right) \), reflecting flips only the first coordinate.
So \( 135^\circ = \tfrac{3\pi}{4} \), and the point is \( \mathbf{\left(-\tfrac{\sqrt{2}}{2},\ \tfrac{\sqrt{2}}{2}\right)} \).
- Start from the Pythagorean identity: \( \sin^2 t + \cos^2 t = 1 \).
- Substitute \( \cos t = \tfrac{3}{5} \): \( \sin^2 t = 1 - \left(\tfrac{3}{5}\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25} \).
- Take the square root: \( \sin t = \pm\tfrac{4}{5} \).
- In Quadrant IV the \( y \)-coordinate is negative, so we choose the minus sign.
Therefore \( \sin t = \mathbf{-\tfrac{4}{5}} \).
Practice
Try each one yourself, then reveal the full solution.
1. Convert \( 60^\circ \) to radians, and state the point \( (\cos 60^\circ, \sin 60^\circ) \) on the unit circle.
Convert by multiplying by \( \tfrac{\pi}{180} \): \( 60 \times \tfrac{\pi}{180} = \tfrac{60\pi}{180} = \tfrac{\pi}{3} \).
The special angle \( 60^\circ \) sits in Quadrant I with coordinates \( \left(\tfrac{1}{2},\ \tfrac{\sqrt{3}}{2}\right) \).
So \( 60^\circ = \tfrac{\pi}{3} \) and the point is \( \mathbf{\left(\tfrac{1}{2},\ \tfrac{\sqrt{3}}{2}\right)} \).
2. In which quadrant is an angle whose sine is positive but whose cosine is negative? Give one example angle.
Sine is the \( y \)-coordinate, so \( \sin t > 0 \) means the point is above the \( x \)-axis: Quadrant I or II.
Cosine is the \( x \)-coordinate, so \( \cos t < 0 \) means the point is left of the \( y \)-axis: Quadrant II or III.
The overlap is Quadrant II — which matches the "S" (Sine positive) in All Students Take Calculus. An example is \( 120^\circ \), whose point is \( \left(-\tfrac{1}{2},\ \tfrac{\sqrt{3}}{2}\right) \).
3. Given \( \sin t = \tfrac{5}{13} \) with \( t \) in Quadrant II, find \( \cos t \).
Use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \).
Substitute \( \sin t = \tfrac{5}{13} \): \( \cos^2 t = 1 - \left(\tfrac{5}{13}\right)^2 = 1 - \tfrac{25}{169} = \tfrac{144}{169} \).
Take the root: \( \cos t = \pm\tfrac{12}{13} \). In Quadrant II the \( x \)-coordinate is negative, so we keep the minus sign.
So \( \cos t = \mathbf{-\tfrac{12}{13}} \).