The unit circle

The right-triangle definitions of sine and cosine work perfectly for acute angles, but they break down once an angle exceeds $90\degree$. To extend trigonometry to any angle we need a new framework, and the unit circle provides it.

The radius being exactly $1$ is what makes this circle special: as we will see later in this lesson, it creates a direct link between a point's coordinates and the trigonometric ratios.

To use the unit circle we place angles in standard position: vertex at the origin, initial side along the positive $x$-axis, measuring anticlockwise. The other ray, the terminal side, sweeps out the angle $\theta$ and intersects the unit circle at a single point called the terminal point of $\theta$.

The four points where the axes cross the unit circle correspond to the quadrantal angles:

Each of these satisfies $x^2 + y^2 = 1$.

An angle of $\theta = 0$ has its terminal side along the positive $x$-axis, so the terminal point is $(1, 0)$. An angle of $\theta = \dfrac{\pi}{2}$ points straight up along the positive $y$-axis, giving the terminal point $(0, 1)$. Verify that both points lie on the unit circle.

Solution

A point lies on the unit circle if it satisfies $x^2 + y^2 = 1$.

For $(1, 0)$:

For $(0, 1)$:

Both points satisfy the equation, confirming they lie on the unit circle.

We start by recalling the right-triangle definitions of cosine and sine:

On the unit circle, these ratios collapse into something much simpler.

To see why, take an angle $\theta$ in standard position, mark its terminal point on the unit circle, and drop a perpendicular from that point to the $x$-axis. This forms a right triangle whose hypotenuse is the radius of the circle - exactly $1$.

Applying the right-triangle definitions to this triangle:

The hypotenuse being exactly $1$ is the key: the ratios collapse to the coordinates themselves.

The terminal point of an angle $\theta$ on the unit circle is $\left(\dfrac{3}{5},\, \dfrac{4}{5}\right)$. State $\cos\theta$ and $\sin\theta$, and verify that the point lies on the unit circle.

Solution

On the unit circle, the terminal point has coordinates $(\cos\theta,\, \sin\theta)$. Reading off the coordinates directly:

To verify the point lies on the unit circle, we check that $x^2 + y^2 = 1$:

The point satisfies $x^2 + y^2 = 1$, confirming it lies on the unit circle.

Visually, this is the famous $3$-$4$-$5$ right triangle inscribed in the unit circle, with each side scaled by $\dfrac{1}{5}$ so the hypotenuse equals $1$. The adjacent side has length $\dfrac{3}{5} = \cos\theta$ and the opposite side has length $\dfrac{4}{5} = \sin\theta$.

When an angle $\theta$ lands outside quadrant I, it helps to relate it back to a familiar acute angle. The reference angle gives us exactly that.

The reference angle is always measured to the $x$-axis, never the $y$-axis. The formula for $\alpha$ depends on which quadrant the terminal side falls in:

If $\theta$ falls outside the range $[0, 2\pi)$, first find a coterminal angle in that range by adding or subtracting $2\pi$, then apply the appropriate formula.

Find the reference angle for $\theta = \dfrac{5\pi}{3}$.

Solution

First, identify the quadrant. Since $\dfrac{3\pi}{2} < \dfrac{5\pi}{3} < 2\pi$, the terminal side is in quadrant IV.

For an angle in quadrant IV, the reference angle formula is $\alpha = 2\pi - \theta$:

The reference angle is $\dfrac{\pi}{3}$, which is the acute angle between the terminal side and the positive $x$-axis.