In geometry, sine and cosine were ratios of sides in a right triangle, so they only made sense for angles between $0^\circ$ and $90^\circ$. Now that angles can be any size, we need a definition that works for $150^\circ$, $-\frac{\pi}{3}$ or $1000^\circ$. The unit circle, the circle of radius $1$ centered at the origin, gives it.
Key idea
Draw $\theta$ in standard position. Its terminal side meets the unit circle at one point, and that point is $$(\cos\theta, \sin\theta).$$ Cosine is the $x$-coordinate and sine is the $y$-coordinate.
Why it works
For an acute angle, drop a line from the point $(x, y)$ straight down to the $x$-axis. You get a right triangle with hypotenuse $1$, adjacent side $x$ and opposite side $y$. So the triangle ratios give $\cos\theta = \frac{x}{1} = x$ and $\sin\theta = \frac{y}{1} = y$. The new definition agrees with the old one where the old one works, and it keeps working when the point moves into other quadrants, where $x$ or $y$ can be negative.
Worked examples
Worked example: Reading a point
The terminal side of $\theta$ meets the unit circle at $\left(-\frac{3}{5}, \frac{4}{5}\right)$. Find $\cos\theta$ and $\sin\theta$.
- The point is $(\cos\theta, \sin\theta)$.
- So $\cos\theta = -\frac{3}{5}$ and $\sin\theta = \frac{4}{5}$. (Check: $\frac{9}{25} + \frac{16}{25} = 1$, so the point really is on the unit circle.)
Answer: $\cos\theta = -\frac{3}{5}$, $\sin\theta = \frac{4}{5}$
Worked example: An angle on an axis
Find $\cos\frac{3\pi}{2}$ and $\sin\frac{3\pi}{2}$.
- $\frac{3\pi}{2}$ is three quarters of a turn, pointing straight down.
- It meets the unit circle at $(0, -1)$.
Answer: $\cos\frac{3\pi}{2} = 0$, $\sin\frac{3\pi}{2} = -1$
Worked example: A bigger circle
The point $(6, -8)$ is on the terminal side of $\theta$. Find $\cos\theta$ and $\sin\theta$.
- Its distance from the origin is $\sqrt{36 + 64} = 10$.
- Shrink by $10$ to reach the unit circle: $(0.6, -0.8)$.
- So $\cos\theta = \frac{3}{5}$ and $\sin\theta = -\frac{4}{5}$.
Answer: $\cos\theta = \frac{3}{5}$, $\sin\theta = -\frac{4}{5}$
Common mistake
Swapping the coordinates and taking sine as the $x$-value. Remember the order: $(\cos, \sin)$ matches $(x, y)$, the same alphabetical order. Quick check: at $90^\circ$ the point is $(0, 1)$, the top of the circle, and $\sin 90^\circ$ should be $1$ because sine measures height.