Welcome to Exact Trigonometric Values!

Hi there! This chapter is incredibly important, especially for your non-calculator papers. Why? Because sometimes, math doesn't want you to have an approximate answer (like 0.707...), it demands the exact value (like \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{3}}{2}\)).

In this section, we will learn how to derive and recall the exact sine, cosine, and tangent values for a few very special angles: \(0^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). These values are cornerstones of trigonometry!

Section 1: Why We Need Exact Values

When you use a calculator for \(\sin(60^{\circ})\), you get approximately 0.866. But the exact value is \(\frac{\sqrt{3}}{2}\). In many exam contexts, especially when working with surds (roots), you must use the exact form.

Key Takeaway: Exact values involve integers and surds (like \(\sqrt{2}\) or \(\sqrt{3}\)) and are essential for questions where no calculator is permitted.

Section 2: Deriving Exact Values Using Special Triangles

We can figure out most of these values using two simple right-angled triangles. Don't worry if this seems tricky at first; drawing the triangles is the perfect revision tool!

2.1 The \(45^{\circ}\) Triangle (The Isosceles Right-Angled Triangle)

Start with a simple square with sides of length 1 unit.

  1. Cut the square diagonally to create two right-angled triangles.
  2. The angles inside are \(90^{\circ}\), \(45^{\circ}\), and \(45^{\circ}\).
  3. The two shorter sides are both length 1.
  4. Use Pythagoras' Theorem (\(a^2 + b^2 = c^2\)) to find the hypotenuse:
    \(c^2 = 1^2 + 1^2 = 2\)
    \(c = \sqrt{2}\)

Using SOH CAH TOA on this triangle:

  • Sine \(45^{\circ}\) (\(SOH\)): Opposite / Hypotenuse
    \(\sin(45^{\circ}) = \frac{1}{\sqrt{2}}\) (Often rationalised to \(\frac{\sqrt{2}}{2}\))
  • Cosine \(45^{\circ}\) (\(CAH\)): Adjacent / Hypotenuse
    \(\cos(45^{\circ}) = \frac{1}{\sqrt{2}}\) (Often rationalised to \(\frac{\sqrt{2}}{2}\))
  • Tangent \(45^{\circ}\) (\(TOA\)): Opposite / Adjacent
    \(\tan(45^{\circ}) = \frac{1}{1} = 1\)

Memory Aid: Since the two non-right angles are equal, the Sine and Cosine values for \(45^{\circ}\) must be the same. And \(\tan(45^{\circ})\) is always 1!

2.2 The \(30^{\circ}\) and \(60^{\circ}\) Triangle (The Half-Equilateral Triangle)

Start with an equilateral triangle (all sides and angles equal). Let the side length be 2 units (this makes the calculations cleaner).

  1. All angles are \(60^{\circ}\). All sides are length 2.
  2. Draw a perpendicular line from the top vertex to the base. This splits the triangle exactly in half, creating a right-angled triangle.
  3. The angles in the new triangle are \(90^{\circ}\), \(60^{\circ}\), and \(30^{\circ}\).
  4. The hypotenuse remains 2. The base is now 1 (half of 2).
  5. Use Pythagoras to find the remaining side (the height, \(h\)):
    \(1^2 + h^2 = 2^2\)
    \(1 + h^2 = 4\)
    \(h^2 = 3\)
    \(h = \sqrt{3}\)

Using SOH CAH TOA on this 1-\(\sqrt{3}\)-2 triangle:

For \(30^{\circ}\):

  • \(\sin(30^{\circ}) = \text{Opposite}/\text{Hypotenuse} = \frac{1}{2}\)
  • \(\cos(30^{\circ}) = \text{Adjacent}/\text{Hypotenuse} = \frac{\sqrt{3}}{2}\)
  • \(\tan(30^{\circ}) = \text{Opposite}/\text{Adjacent} = \frac{1}{\sqrt{3}}\) (Often rationalised to \(\frac{\sqrt{3}}{3}\))

For \(60^{\circ}\):

  • \(\sin(60^{\circ}) = \text{Opposite}/\text{Hypotenuse} = \frac{\sqrt{3}}{2}\)
  • \(\cos(60^{\circ}) = \text{Adjacent}/\text{Hypotenuse} = \frac{1}{2}\)
  • \(\tan(60^{\circ}) = \text{Opposite}/\text{Adjacent} = \frac{\sqrt{3}}{1} = \sqrt{3}\)

Did You Know? Notice that \(\sin(30^{\circ})\) is the same as \(\cos(60^{\circ})\), and \(\cos(30^{\circ})\) is the same as \(\sin(60^{\circ})\). This is because \(30^{\circ} + 60^{\circ} = 90^{\circ}\)!

Quick Review: The Two Essential Triangles

If you forget the values, draw these instantly:

Triangle 1 (45°): Sides 1, 1, \(\sqrt{2}\).
Triangle 2 (30°, 60°): Sides 1, \(\sqrt{3}\), 2 (Hypotenuse is 2).

Section 3: The Extremes: \(0^{\circ}\) and \(90^{\circ}\)

The values for \(0^{\circ}\) and \(90^{\circ}\) are slightly different because they don't form a "proper" triangle (they are limiting cases). We can understand these by thinking about a right-angled triangle where one angle gets extremely small (\(0^{\circ}\)) or extremely large (\(90^{\circ}\)).

Imagine a tiny angle (\(\theta\)) in a right-angled triangle:

  • As \(\theta \to 0^{\circ}\), the Opposite side gets tiny (close to 0), and the Adjacent side gets almost as long as the Hypotenuse (close to 1).
  • As \(\theta \to 90^{\circ}\), the Opposite side gets almost as long as the Hypotenuse (close to 1), and the Adjacent side gets tiny (close to 0).

The Exact Values for \(0^{\circ}\) and \(90^{\circ}\):

  • \(\sin(0^{\circ}) = 0\),   \(\cos(90^{\circ}) = 0\)
  • \(\cos(0^{\circ}) = 1\),   \(\sin(90^{\circ}) = 1\)
  • \(\tan(0^{\circ}) = \frac{\sin(0^{\circ})}{\cos(0^{\circ})} = \frac{0}{1} = 0\)
  • \(\tan(90^{\circ})\): This is \(\frac{\sin(90^{\circ})}{\cos(90^{\circ})} = \frac{1}{0}\). Division by zero is undefined. We call this undefined or infinite.

Section 4: The Ultimate Memory Trick (The Square Root Method)

If you don't have time to draw triangles during an exam, this method is the fastest way to reconstruct the entire table for Sine and Cosine.

Step-by-Step Guide for Sine and Cosine (0°, 30°, 45°, 60°, 90°)
  1. Write the Angle Order: Write the angles from \(0^{\circ}\) to \(90^{\circ}\).
  2. Write the Index Numbers:
    • For Sine, write the sequence: 0, 1, 2, 3, 4.
    • For Cosine, write the sequence in reverse: 4, 3, 2, 1, 0.
  3. Apply the Formula: The exact value is always: \(\frac{\sqrt{\text{Index Number}}}{2}\)

Example: Finding \(\sin(60^{\circ})\)

\(60^{\circ}\) is the third index number in the Sine sequence (0, 1, 2, 3, 4).
Value = \(\frac{\sqrt{3}}{2}\). (Matches our triangle derivation!)

Example: Finding \(\cos(45^{\circ})\)

\(45^{\circ}\) has index number 2 in the Cosine sequence (4, 3, 2, 1, 0).
Value = \(\frac{\sqrt{2}}{2}\). (Matches our triangle derivation!)

Don't Forget Tangent: Once you have Sine and Cosine, remember the identity:
\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)

Common Mistake to Avoid

When you find \(\sin(30^{\circ}) = \frac{\sqrt{1}}{2}\), remember that \(\sqrt{1} = 1\). So the answer is simply \(\frac{1}{2}\).
Similarly, \(\sin(0^{\circ}) = \frac{\sqrt{0}}{2} = 0\).

Section 5: The Master Table of Exact Values

You need to memorize or be able to quickly generate the values in this table.

Angle (\(x\)) \(0^{\circ}\) \(30^{\circ}\) \(45^{\circ}\) \(60^{\circ}\) \(90^{\circ}\)
\(\sin(x)\) \(0\) \(\frac{1}{2}\) \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) \(\frac{\sqrt{3}}{2}\) \(1\)
\(\cos(x)\) \(1\) \(\frac{\sqrt{3}}{2}\) \(\frac{\sqrt{2}}{2}\) or \(\frac{1}{\sqrt{2}}\) \(\frac{1}{2}\) \(0\)
\(\tan(x)\) \(0\) \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\) \(1\) \(\sqrt{3}\) Undefined


Key Takeaway: Master the pattern (0, 1, 2, 3, 4) for the numerators of the square roots. This ensures you can rebuild the entire table in under 30 seconds when you start your exam!

You've got this! Knowing these exact values is a guaranteed way to score marks in non-calculator trigonometry questions. Practice drawing those two special triangles until they are second nature.