Mathematics (0580) | Exact trigonometric values

Mastering Exact Trigonometric Values (0580 Extended)

Welcome to one of the most crucial topics for your non-calculator trigonometry questions! When we talk about exact trigonometric values, we are learning how to find the Sine, Cosine, and Tangent of specific special angles (like 30°, 45°, and 60°) without using a calculator and giving the answer perfectly—not as rounded decimals.

Why is this important?
In your exam, especially the non-calculator paper (Paper 2), you must provide the answer in its exact form (usually involving fractions or surds, like \(\frac{\sqrt{3}}{2}\)). If you give a decimal approximation (like 0.866), you will lose marks! Mastering these values is like having a secret construction kit for solving complex geometry problems.

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1. What are Exact Values?

When you type \(\sin(60^\circ)\) into a calculator, it might show 0.8660254... This is a decimal approximation. The true, exact value is \(\frac{\sqrt{3}}{2}\).

An Exact Value is a value expressed using integers (whole numbers), fractions, and/or surds (roots that can't be simplified to a whole number, like \(\sqrt{2}\) or \(\sqrt{3}\)).

Key Rule: Always use Surds and Fractions

If a question asks for the exact value of \(\sin(45^\circ)\), you must write \(\frac{1}{\sqrt{2}}\) or the rationalised form \(\frac{\sqrt{2}}{2}\). You must never write 0.707.

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2. The Source: Building the Special Right Triangles

All exact trigonometric values come from just two simple right-angled triangles. If you can draw and label these two triangles, you can derive all the exact values you need.

2.1. The 45° Triangle (Isosceles Right Triangle)

This triangle comes from cutting a perfect square in half diagonally.

Step 1: Start with a Square

• Imagine a square with sides of length 1 unit.
• Cut it diagonally to form two right-angled triangles.
• Since the original angles were 90°, cutting through the corners creates two 45° angles.

Step 2: Find the Hypotenuse

Using Pythagoras' Theorem (\(a^2 + b^2 = c^2\)):

\(1^2 + 1^2 = c^2\)
\(1 + 1 = c^2\)
\(c^2 = 2\)
\(c = \sqrt{2}\)

The triangle sides are 1, 1, and \(\sqrt{2}\).

Step 3: Calculate the Values for 45° (SOH CAH TOA)

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

2.2. The 30° and 60° Triangle (Half-Equilateral Triangle)

This triangle comes from cutting an equilateral triangle in half.

Step 1: Start with an Equilateral Triangle

• Imagine an equilateral triangle with sides of length 2 units (using 2 simplifies the fractions later).
• All angles are 60°.
• Draw a line straight down the middle (the height). This line cuts the base (2) into two halves (1 and 1) and cuts the top angle (60°) into two 30° angles.

Step 2: Find the Height (Opposite side to 60°)

We now have a right-angled triangle with sides 2 (hypotenuse) and 1 (base). Let \(h\) be the height.

Using Pythagoras' Theorem:

\(1^2 + h^2 = 2^2\)
\(1 + h^2 = 4\)
\(h^2 = 3\)
\(h = \sqrt{3}\)

The triangle sides are 1, \(\sqrt{3}\), and 2 (the hypotenuse). The angles are 30°, 60°, and 90°.

Step 3: Calculate the Values for 30° and 60°

For 30°:

(Relative to the 30° angle, Opposite = 1, Adjacent = \(\sqrt{3}\), Hypotenuse = 2)

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

For 60°:

(Relative to the 60° angle, Opposite = \(\sqrt{3}\), Adjacent = 1, Hypotenuse = 2)

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

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3. Quadrantal Angles (0° and 90°)

These angles are special because they aren't part of a standard triangle, but we can imagine a triangle "collapsing" or "stretching" to figure out their values. Think about the side lengths changing.

3.1. Values for 0°

Imagine a right triangle where the angle is almost 0°.

• The Opposite side becomes 0.
• The Adjacent side becomes equal to the Hypotenuse (let's say 1).

The ratios become:

• \(\sin(0^\circ)\) = Opp/Hyp = \(\frac{0}{1} = 0\)
• \(\cos(0^\circ)\) = Adj/Hyp = \(\frac{1}{1} = 1\)
• \(\tan(0^\circ)\) = Opp/Adj = \(\frac{0}{1} = 0\)

3.2. Values for 90°

Imagine a right triangle where the angle is almost 90°.

• The Opposite side becomes equal to the Hypotenuse (1).
• The Adjacent side becomes 0.

The ratios become:

• \(\sin(90^\circ)\) = Opp/Hyp = \(\frac{1}{1} = 1\)
• \(\cos(90^\circ)\) = Adj/Hyp = \(\frac{0}{1} = 0\)
• \(\tan(90^\circ)\) = Opp/Adj = \(\frac{1}{0}\). This is Undefined, because division by zero is not possible.

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4. Comprehensive Summary Table and Memory Aids

While drawing the triangles is the most reliable method, students often need to recall these values quickly. Here is the complete table and an excellent memory trick.

4.1. The Complete Table of Exact Values

The key to spotting the pattern is writing all the values with a denominator of 2.

\(\theta\)	0°	30°	45°	60°	90°
\(\sin(\theta)\)	0	\(\frac{1}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{3}}{2}\)	1
\(\cos(\theta)\)	1	\(\frac{\sqrt{3}}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{1}{2}\)	0
\(\tan(\theta)\)	0	\(\frac{1}{\sqrt{3}}\)	1	\(\sqrt{3}\)	Undefined

4.2. Memory Trick: The Hand Method

This is an incredibly helpful trick for recalling the Sin and Cos values quickly.

1. Hold your left hand flat, palm facing you. Assign the five key angles (0°, 30°, 45°, 60°, 90°) to your fingers, starting from the thumb (0°) to the pinky (90°).

2. The formula to find the value is: \(\frac{\sqrt{\text{N}}}{2}\) where N is the number of fingers.

3. For Sine: Count the number of fingers below the angle you are testing.

• Example: 30° (Index Finger). There is 1 finger below it (the thumb). So, \(N=1\).
  \(\sin(30^\circ) = \frac{\sqrt{1}}{2} = \frac{1}{2}\)
• Example: 90° (Pinky Finger). There are 4 fingers below it. So, \(N=4\).
  \(\sin(90^\circ) = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1\)

4. For Cosine: Count the number of fingers above the angle you are testing.

• Example: 60° (Ring Finger). There is 1 finger above it (the pinky). So, \(N=1\).
  \(\cos(60^\circ) = \frac{\sqrt{1}}{2} = \frac{1}{2}\)
• Example: 0° (Thumb). There are 4 fingers above it. So, \(N=4\).
  \(\cos(0^\circ) = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1\)

Once you know the Sin and Cos values, you can always find Tangent using the identity: \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)

Key Takeaway

The exact trigonometric values for 0°, 30°, 45°, 60°, and 90° are non-negotiable knowledge for the Extended curriculum. Practice sketching the 45° triangle (1, 1, \(\sqrt{2}\)) and the 30°/60° triangle (1, \(\sqrt{3}\), 2) until you can draw them perfectly—they are your safety net if you forget the table or the hand trick!