Unit 1: Pure Mathematics — Trigonometric Equations
Welcome to the study notes for Trigonometric Equations! In regular GCSE Mathematics, you used trigonometry to find missing sides and angles in right-angled and non-right-angled triangles. In GCSE Further Mathematics, we take trigonometry a step further by treating trigonometric ratios as repeating wave functions and solving equations involving them.
Don't worry if this seems a bit daunting at first! Once you learn a few standard methods and a couple of golden identities, solving these equations becomes just like solving regular algebraic equations, with one extra step: finding all the matching angles in a given range.
Did you know? Trigonometric functions model repeating cycles in the real world — from sound waves and light waves to tidal patterns, alternating electrical currents (AC), and the rotation of Ferris wheels!
1. Prerequisite Knowledge & Fundamental Identities
Before solving equations, let's review two fundamental trigonometric identities that you must know inside out for CCEA Further Mathematics:
Identity 1: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
Identity 2: \(\sin^2 \theta + \cos^2 \theta = 1\) (which can also be rearranged as \(\sin^2 \theta = 1 - \cos^2 \theta\) or \(\cos^2 \theta = 1 - \sin^2 \theta\))
Note on notation: \(\sin^2 \theta\) means \((\sin \theta)^2\). It does not mean \(\sin(\theta^2)\).
Quick Review of Exact Values
While you can use your calculator, knowing standard exact values for \(0^\circ, 30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\) helps you spot patterns quickly:
• \(\sin 30^\circ = \frac{1}{2}\), \(\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
• \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\), \(\cos 60^\circ = \frac{1}{2}\)
• \(\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\), \(\tan 45^\circ = 1\), \(\tan 60^\circ = \sqrt{3}\)
Key Takeaway: Whenever you see \(\sin^2 \theta\) and \(\cos \theta\) mixed in an equation, use \(\sin^2 \theta = 1 - \cos^2 \theta\) so the whole equation is written in terms of a single trigonometric function.
2. Understanding Why There Are Multiple Solutions
When you solve an equation like \(\sin \theta = 0.5\), typing \(\sin^{-1}(0.5)\) into your calculator gives you \(30^\circ\). This is called the principal value (or acute reference angle).
However, because trigonometric graphs repeat and have symmetry, there are other angles in the range \(0^\circ \le \theta \le 360^\circ\) that have the exact same sine value. For instance, \(\sin 150^\circ = 0.5\) as well!
Methods to Find All Solutions
You can use either the graphical symmetry method or the CAST diagram (quadrant rule). Both methods give the same correct answers.
Method A: Graph Symmetry Rules (for \(0^\circ \le \theta \le 360^\circ\))
• For Sine (\(\sin \theta\)): If \(\alpha\) is your first solution, the second is \(180^\circ - \alpha\).
• For Cosine (\(\cos \theta\)): If \(\alpha\) is your first solution, the second is \(360^\circ - \alpha\).
• For Tangent (\(\tan \theta\)): If \(\alpha\) is your first solution, the next is \(\alpha + 180^\circ\) (since the tangent wave repeats every \(180^\circ\)).
Method B: The CAST Diagram (Quadrant System)
The quadrants go anti-clockwise from \(0^\circ\) to \(360^\circ\):
• Quadrant 1 (\(0^\circ\) to \(90^\circ\)): All ratios are positive (\(\theta = \alpha\)).
• Quadrant 2 (\(90^\circ\) to \(180^\circ\)): Sin is positive (\(\theta = 180^\circ - \alpha\)).
• Quadrant 3 (\(180^\circ\) to \(270^\circ\)): Tan is positive (\(\theta = 180^\circ + \alpha\)).
• Quadrant 4 (\(270^\circ\) to \(360^\circ\)): Cos is positive (\(\theta = 360^\circ - \alpha\)).
Memory Trick: Remember CAST starting from Quadrant 4 and moving anti-clockwise: Cos, All, Sin, Tan (or "Cats Are So Tame").
Key Takeaway: Your calculator gives only one solution. Always use symmetry or the CAST diagram to find all other valid solutions in the required range.
3. Type 1: Basic Linear Trigonometric Equations
These equations have the form \(a \sin \theta + b = 0\), \(a \cos \theta + b = 0\), or \(a \tan \theta + b = 0\).
Step-by-Step Strategy:
1. Rearrange the equation to isolate the trigonometric function (e.g., \(\sin \theta = c\)).
2. Find the reference angle \(\alpha = \sin^{-1}(|c|)\) using positive values to avoid sign confusion.
3. Determine which quadrants your solutions lie in based on whether \(c\) is positive or negative.
4. Calculate the angles in the given range.
Worked Example 1
Solve \(2\cos \theta + 1 = 0\) for \(0^\circ \le \theta \le 360^\circ\).
Step 1: Rearrange to isolate \(\cos \theta\):
\(2\cos \theta = -1\)
\(\cos \theta = -0.5\)
Step 2: Find the basic acute angle:
\(\alpha = \cos^{-1}(0.5) = 60^\circ\)
Step 3: Identify the quadrants:
Since \(\cos \theta\) is negative, solutions lie in Quadrant 2 and Quadrant 3.
Step 4: Calculate the solutions:
Quadrant 2: \(\theta = 180^\circ - 60^\circ = 120^\circ\)
Quadrant 3: \(\theta = 180^\circ + 60^\circ = 240^\circ\)
Final Answer: \(\theta = 120^\circ, 240^\circ\)
Key Takeaway: Always check your answers at the end by substituting them back into the original equation on your calculator (e.g., \(2\cos(120^\circ) + 1 = 0\)).
4. Type 2: Equations Involving Multiple Angles & Phase Shifts
Sometimes the angle inside the function is modified, such as \(\sin(2\theta) = 0.5\) or \(\cos(\theta - 45^\circ) = \frac{\sqrt{3}}{2}\).
Step-by-Step Strategy:
1. Adjust the interval to match the angle inside the function.
2. Solve for the combined angle first (e.g., solve for \(2\theta\) or \((\theta - 45^\circ)\)).
3. Finally, rearrange to find \(\theta\).
4. Discard any values outside the original range for \(\theta\).
Worked Example 2: Multiple Angle
Solve \(\sin(2\theta) = \frac{\sqrt{3}}{2}\) for the interval \(0^\circ \le \theta \le 360^\circ\).
Step 1: Adjust the range:
Since \(0^\circ \le \theta \le 360^\circ\), multiplying by \(2\) gives the new range for \(2\theta\):
\(0^\circ \le 2\theta \le 720^\circ\)
Step 2: Find reference angle and solutions for \(2\theta\):
\(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ\)
Sine is positive in Quadrants 1 and 2.
First cycle (\(0^\circ\) to \(360^\circ\)):
\(2\theta = 60^\circ\) and \(2\theta = 180^\circ - 60^\circ = 120^\circ\)
Second cycle (\(360^\circ\) to \(720^\circ\)) — add \(360^\circ\) to each solution:
\(2\theta = 60^\circ + 360^\circ = 420^\circ\)
\(2\theta = 120^\circ + 360^\circ = 480^\circ\)
Step 3: Divide every value by 2 to solve for \(\theta\):
\(\theta = \frac{60^\circ}{2} = 30^\circ\)
\(\theta = \frac{120^\circ}{2} = 60^\circ\)
\(\theta = \frac{420^\circ}{2} = 210^\circ\)
\(\theta = \frac{480^\circ}{2} = 240^\circ\)
Final Answer: \(\theta = 30^\circ, 60^\circ, 210^\circ, 240^\circ\)
Key Takeaway: Never divide by the multiplier early! Always find all solutions for the transformed angle (e.g., \(2\theta\)) across the extended range before dividing by \(2\).
5. Type 3: Equations Involving \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
When an equation has \(\sin \theta\) on one side and \(\cos \theta\) on the other, you can often divide both sides by \(\cos \theta\) to create a single \(\tan \theta\) equation.
Worked Example 3
Solve \(3\sin \theta = 5\cos \theta\) for \(0^\circ \le \theta \le 360^\circ\).
Step 1: Divide both sides by \(\cos \theta\):
\(3\left(\frac{\sin \theta}{\cos \theta}\right) = 5\)
\(3\tan \theta = 5\)
Step 2: Isolate \(\tan \theta\):
\(\tan \theta = \frac{5}{3} \approx 1.6667\)
Step 3: Find the angles:
\(\alpha = \tan^{-1}\left(\frac{5}{3}\right) \approx 59.04^\circ\)
Since \(\tan \theta\) is positive, solutions are in Quadrants 1 and 3:
Quadrant 1: \(\theta = 59.04^\circ\)
Quadrant 3: \(\theta = 59.04^\circ + 180^\circ = 239.04^\circ\)
Final Answer: \(\theta = 59.0^\circ, 239.0^\circ\) (to 1 d.p.)
Key Takeaway: When \(\sin \theta\) and \(\cos \theta\) appear with no constant term, dividing by \(\cos \theta\) simplifies the equation into a single \(\tan \theta\) term.
6. Type 4: Quadratic Trigonometric Equations
These equations look like algebraic quadratics: \(a x^2 + b x + c = 0\), but where \(x\) is replaced by \(\sin \theta\), \(\cos \theta\), or \(\tan \theta\).
Worked Example 4: Direct Quadratic
Solve \(2\sin^2 \theta - \sin \theta - 1 = 0\) for \(0^\circ \le \theta \le 360^\circ\).
Step 1: Factorise the quadratic expression:
Let \(s = \sin \theta\). Then \(2s^2 - s - 1 = 0\).
\((2s + 1)(s - 1) = 0\)
So: \((2\sin \theta + 1)(\sin \theta - 1) = 0\)
Step 2: Solve each bracket separately:
Case 1: \(2\sin \theta + 1 = 0 \implies \sin \theta = -0.5\)
Reference angle: \(\sin^{-1}(0.5) = 30^\circ\)
Since \(\sin \theta\) is negative (Quadrants 3 & 4):
\(\theta = 180^\circ + 30^\circ = 210^\circ\)
\(\theta = 360^\circ - 30^\circ = 330^\circ\)
Case 2: \(\sin \theta - 1 = 0 \implies \sin \theta = 1\)
\(\theta = 90^\circ\)
Step 3: List all valid solutions:
Final Answer: \(\theta = 90^\circ, 210^\circ, 330^\circ\)
Worked Example 5: Quadratic Requiring Identity Substitution
Solve \(2\cos^2 \theta + 3\sin \theta = 3\) for \(0^\circ \le \theta \le 360^\circ\).
Step 1: Use \(\cos^2 \theta = 1 - \sin^2 \theta\) to get only sine terms:
\(2(1 - \sin^2 \theta) + 3\sin \theta = 3\)
\(2 - 2\sin^2 \theta + 3\sin \theta = 3\)
Step 2: Rearrange to standard quadratic form:
\(0 = 2\sin^2 \theta - 3\sin \theta + 1\)
\((2\sin \theta - 1)(\sin \theta - 1) = 0\)
Step 3: Solve each branch:
• \(2\sin \theta - 1 = 0 \implies \sin \theta = 0.5 \implies \theta = 30^\circ, 150^\circ\)
• \(\sin \theta - 1 = 0 \implies \sin \theta = 1 \implies \theta = 90^\circ\)
Final Answer: \(\theta = 30^\circ, 90^\circ, 150^\circ\)
Key Takeaway: If an equation contains both \(\cos^2 \theta\) and \(\sin \theta\), rewrite the squared term using \(\sin^2 \theta + \cos^2 \theta = 1\) so that the entire equation is in terms of one single trig ratio.
7. Common Pitfalls to Avoid
• Cancelling trigonometric terms: Never divide both sides of an equation by \(\sin \theta\) or \(\cos \theta\) if it causes you to lose solutions. For example, with \(\sin^2 \theta = \sin \theta\), factorise as \(\sin \theta(\sin \theta - 1) = 0\). If you divide by \(\sin \theta\), you will lose the solutions where \(\sin \theta = 0\).
• Wrong Calculator Mode: In CCEA GCSE Further Mathematics Unit 1, angles are predominantly in degrees unless specified otherwise. Make sure your calculator displays D (Degrees), not R (Radians).
• Forgetting Range Limits: Pay close attention to inequality signs like \(0^\circ \le \theta \le 360^\circ\) vs \(0^\circ < \theta < 360^\circ\). If \(0^\circ\) is a solution, only include it if \(\le\) is used.
• Invalid Solutions: Remember that \(-1 \le \sin \theta \le 1\) and \(-1 \le \cos \theta \le 1\). If factorising gives \(\sin \theta = 2.5\), state clearly that this yields no real solutions.
8. Quick Review Checklist
• Identity 1: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
• Identity 2: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Sine symmetry: \(\theta\) and \(180^\circ - \theta\)
• Cosine symmetry: \(\theta\) and \(360^\circ - \theta\)
• Tangent symmetry: \(\theta\) and \(180^\circ + \theta\)
• Multiple angles (e.g., \(2\theta\)): Expand interval first, solve for \(2\theta\), then divide by \(2\) last.