Introduction to the Unit Circle

Welcome to one of the most important chapters in IB Mathematics! If you have ever felt like trigonometry is just a collection of random button-presses on a calculator, the Unit Circle is about to change that. It is the "GPS map" of trigonometry, showing us how angles, coordinates, and ratios all fit together in one beautiful circle. Whether you are in SL or HL, mastering this circle is the secret to success in Paper 1, where calculators aren't allowed!

What is the Unit Circle?

The Unit Circle is a circle with a radius of exactly 1 unit, centered at the origin \((0, 0)\) on a coordinate grid. Its equation is \(x^2 + y^2 = 1\).

When we draw an angle \(\theta\) (theta) starting from the positive \(x\)-axis and moving counter-clockwise, the line "hits" the circle at a point \(P(x, y)\). In the world of the unit circle:

  • The x-coordinate is the cosine: \(x = \cos \theta\)
  • The y-coordinate is the sine: \(y = \sin \theta\)
  • The gradient (slope) of the line is the tangent: \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}\)

Quick Review: Remember that in the IB, we use both degrees and radians. You should be comfortable knowing that \(180^\circ = \pi\) radians. Most unit circle questions will use \(\pi\)!

The Four Quadrants and the "CAST" Diagram

As the point \(P\) moves around the circle, the values of \(x\) and \(y\) (and therefore \(\sin\), \(\cos\), and \(\tan\)) change from positive to negative depending on the quadrant.

Quadrant I (\(0\) to \(\frac{\pi}{2}\)): Both \(x\) and \(y\) are positive. All ratios are positive.

Quadrant II (\(\frac{\pi}{2}\) to \(\pi\)): \(x\) is negative, \(y\) is positive. Only Sine is positive.

Quadrant III (\(\pi\) to \(\frac{3\pi}{2}\)): Both \(x\) and \(y\) are negative. Only Tangent is positive (since \(\frac{-y}{-x}\) becomes positive).

Quadrant IV (\(\frac{3\pi}{2}\) to \(2\pi\)): \(x\) is positive, \(y\) is negative. Only Cosine is positive.

Memory Aid: Use the acronym CAST (starting from Quadrant IV and going around) or ASTC (starting from Quadrant I): "All Students Take Chemistry" or "Add Sugar To Coffee".

  • All (Q1)
  • Sine (Q2)
  • Tan (Q3)
  • Cos (Q4)

Exact Trigonometric Values

For Paper 1, you must memorize the exact values for specific "special" angles. These come from two special triangles: the \(45-45-90\) (isosceles right-angled) and the \(30-60-90\) (half an equilateral triangle).

1. The "Must-Know" Values

For \(0\) (\(0^\circ\)):
\(\sin 0 = 0\), \(\cos 0 = 1\), \(\tan 0 = 0\)

For \(\frac{\pi}{6}\) (\(30^\circ\)):
\(\sin \frac{\pi}{6} = \frac{1}{2}\), \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\), \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\) (or \(\frac{\sqrt{3}}{3}\))

For \(\frac{\pi}{4}\) (\(45^\circ\)):
\(\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\) (or \(\frac{\sqrt{2}}{2}\)), \(\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\), \(\tan \frac{\pi}{4} = 1\)

For \(\frac{\pi}{3}\) (\(60^\circ\)):
\(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\), \(\cos \frac{\pi}{3} = \frac{1}{2}\), \(\tan \frac{\pi}{3} = \sqrt{3}\)

For \(\frac{\pi}{2}\) (\(90^\circ\)):
\(\sin \frac{\pi}{2} = 1\), \(\cos \frac{\pi}{2} = 0\), \(\tan \frac{\pi}{2} = \text{undefined}\)

Pro-Tip: Notice a pattern? For Sine, the values are \(\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\). Cosine is just this list in reverse!

Using Reference Angles

What if you are asked for \(\cos \frac{2\pi}{3}\)? Don't panic! Use the Reference Angle (also called the related acute angle).

  1. Find the Quadrant: \(\frac{2\pi}{3}\) is in Quadrant II.
  2. Find the Reference Angle: In Q2, the distance to the horizontal axis is \(\pi - \frac{2\pi}{3} = \frac{\pi}{3}\).
  3. Apply the Sign: In Q2, Cosine is negative.
  4. Combine: \(\cos \frac{2\pi}{3} = -\cos \frac{\pi}{3} = -\frac{1}{2}\).

Key Takeaway: Always relate the angle back to the \(x\)-axis (\(0, \pi,\) or \(2\pi\)) to find your reference angle.

The Ambiguous Case of the Sine Rule

The unit circle helps us understand why the Sine Rule sometimes gives two possible answers. This is known as the Ambiguous Case (SSA).

Because \(\sin \theta = y\), there are two different positions on the unit circle with the same positive \(y\)-value: one in Quadrant I (acute angle \(\theta\)) and one in Quadrant II (obtuse angle \(180^\circ - \theta\)).

Example: If \(\sin \theta = 0.5\), \(\theta\) could be \(30^\circ\) or \(150^\circ\). When solving a triangle, if the side opposite the given angle is shorter than the other given side but longer than the altitude, two distinct triangles can exist!

Common Mistakes to Avoid

  • Mixing up \(x\) and \(y\): Always remember Comes before S alphabetically, just like X comes before Y. So, \((\cos \theta, \sin \theta)\).
  • Calculator Mode: On Paper 2, students often forget to check if their GDC is in Radians or Degrees. Always check this first!
  • Tangent at \(\frac{\pi}{2}\): Remember that \(\tan \theta = \frac{\sin}{\cos}\). At \(\frac{\pi}{2}\), \(\cos\) is \(0\). You cannot divide by zero, so \(\tan \frac{\pi}{2}\) is undefined (this creates an asymptote on a graph).
  • Incorrect Reference Angles: Always measure your reference angle to the horizontal \(x\)-axis, never the vertical \(y\)-axis.

Quick Review Quiz

Try to answer these before checking the notes above!

  1. What is the exact value of \(\sin \frac{\pi}{4}\)?
  2. In which quadrants is \(\tan \theta\) positive?
  3. If \(\cos \theta = -\frac{\sqrt{3}}{2}\) and \(\theta\) is in Q3, what is \(\theta\)?

Answers: 1. \(\frac{\sqrt{2}}{2}\); 2. Quadrants I and III; 3. \(\frac{7\pi}{6}\) (Reference angle is \(\frac{\pi}{6}\), and in Q3 we do \(\pi + \frac{\pi}{6}\)).

Note: For further exploration of how these values create waves, see the chapter on "Circular functions and their graphs". For more complex identities, see "Trigonometric identities and double angle formulae".