Sine and cosine on the unit circle

Beyond the right triangle

In Class 10, $\sin \theta$ was a ratio of two sides of a right triangle. That works only for an acute angle, between $0$ and $\dfrac{\pi}{2}$. But a wheel turns through $3\pi$, a clock hand goes round many times, and an angle is negative when it turns the other way. We need a definition that works for every angle.

The unit circle

Draw a circle with its centre $O$ at the origin and radius $1$. This is the unit circle. Start at the point $A(1, 0)$ and move along the circle a distance $x$. Move anticlockwise when $x$ is positive and clockwise when $x$ is negative. You stop at a point $P$.

The radius is $1$, so the arc length $x$ is also the angle $AOP$ in radians.

Definition. If the point reached is $P = (a, b)$, then

• $\cos x = a$, the x-coordinate of $P$;
• $\sin x = b$, the y-coordinate of $P$.

For an acute angle this agrees with the triangle. Drop a perpendicular from $P$ to the x-axis. The hypotenuse $OP$ is $1$, so $\sin x = \dfrac{b}{1} = b$ and $\cos x = \dfrac{a}{1} = a$.

Every real number $x$ gives exactly one point $P$. So sine and cosine are now functions of every real number.

Push the slider past $6.28$, which is about $2\pi$. The point starts a second turn, and the readouts repeat.

The first identity

The point $P$ lies on the circle, so its distance from $O$ is $1$, and $a^2 + b^2 = 1$. In terms of $x$:

Here $\sin^2 x$ means $(\sin x)^2$, the square of the value. No coordinate of a point on this circle can be more than $1$ in size, so

Four quarter turns

After $2\pi$ the point is back at $A$. So, with $n$ any integer:

• $\sin x = 0$ exactly when $x = n\pi$;
• $\cos x = 0$ exactly when $x = (2n + 1)\dfrac{\pi}{2}$, an odd multiple of $\dfrac{\pi}{2}$.

Four more functions

From sine and cosine we build four more functions. Each one is left out where its denominator is zero.

• $\tan x = \dfrac{\sin x}{\cos x}$, for $x \ne (2n + 1)\dfrac{\pi}{2}$;
• $\cot x = \dfrac{\cos x}{\sin x}$, for $x \ne n\pi$;
• $\sec x = \dfrac{1}{\cos x}$, for $x \ne (2n + 1)\dfrac{\pi}{2}$;
• $\operatorname{cosec} x = \dfrac{1}{\sin x}$, for $x \ne n\pi$.

Divide $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$, and then by $\sin^2 x$. Two more identities follow:

A worked example

We know that $\cos x = -\dfrac{12}{13}$ and that $x$ lies between $\pi$ and $\dfrac{3\pi}{2}$. Let us find the other five functions.

Step 1. The size of $\sin x$. $\sin^2 x = 1 - \dfrac{144}{169} = \dfrac{25}{169}$, so $\sin x = \pm\dfrac{5}{13}$.

Step 2. The sign. For $x$ between $\pi$ and $\dfrac{3\pi}{2}$, the point $P$ is below the x-axis. So $\sin x = -\dfrac{5}{13}$.

Step 3. The rest. $\tan x = \dfrac{-5/13}{-12/13} = \dfrac{5}{12}$ and $\cot x = \dfrac{12}{5}$. Also $\sec x = -\dfrac{13}{12}$ and $\operatorname{cosec} x = -\dfrac{13}{5}$.

Practise with new numbers

Try it in Python

This program finds twelve points on the unit circle, at steps of $\dfrac{\pi}{6}$. It prints $\cos x$ and $\sin x$ for each one, checks the identity, and draws the points.

Common mistakes

• Reading $\sin^2 x$ as $\sin(x^2)$. It means $(\sin x)^2$, the square of the value of the sine.
• Forgetting the sign. $\sin^2 x = \dfrac{25}{169}$ gives two values. The position of $P$ decides which one is right.
• Using a triangle for a large angle. No right triangle has an angle of $\dfrac{5\pi}{6}$. Use the point on the circle.
• Dividing by zero. $\tan \dfrac{\pi}{2}$ is not defined, because $\cos \dfrac{\pi}{2} = 0$.

Remember these

• On the unit circle, the point at angle $x$ is $(\cos x, \sin x)$.
• $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \operatorname{cosec}^2 x$.
• $\sin x = 0$ at $x = n\pi$, and $\cos x = 0$ at the odd multiples of $\dfrac{\pi}{2}$.
• Both sine and cosine lie between $-1$ and $1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The point $P$ on the unit circle at angle $x$ is $\left(\tfrac{8}{17}, -\tfrac{15}{17}\right)$. What are $\sin x$ and $\tan x$?

2. Can $\sin x$ be equal to $1.2$ for some angle $x$? Why?

3. For which $x$ between $0$ and $2\pi$ is $\cos x = 0$?

4. Show that $\sec^2 x - \tan^2 x = 1$ wherever both are defined.

5. Why is $\cot \pi$ not defined, while $\cot \dfrac{\pi}{2} = 0$?