👋 Welcome to FP1 Trigonometry: Finding Infinite Solutions!
Hello! Trigonometry is more than just right-angled triangles—it’s about waves, cycles, and infinite possibilities. In AS Pure Maths, you learned how to find solutions in a specific range (like \(0^\circ\) to \(360^\circ\) or \(0\) to \(2\pi\)).
In Further Maths, we go one step further: we learn how to find the General Solution. This means finding a single mathematical expression that represents every single possible answer for a trigonometric equation, no matter how many times the wave repeats.
Don't worry if this sounds complex! We'll break down the specific rules for sine, cosine, and tangent. Mastering these rules is key to tackling the trickier equations you'll face in FP1.
Key Takeaway from the Introduction
The core focus of FP1 Trigonometry is finding expressions for the infinite number of solutions to an equation, known as the General Solution.
1. Prerequisite Review: Exact Values and Radians
Before diving into general solutions, you must be absolutely comfortable using radians and knowing the exact values for key angles. The syllabus requires you to use these values without relying on a calculator (unless the value is non-exact, like 0.3 or -0.2).
Radian Equivalents for Key Angles
The syllabus specifically mentions these three angles:
- \(30^\circ\) is \(\frac{\pi}{6}\) radians.
- \(45^\circ\) is \(\frac{\pi}{4}\) radians.
- \(60^\circ\) is \(\frac{\pi}{3}\) radians.
Memory Aid (The Exact Values Table):
A handy trick is to think of the sine values as \(\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}\) for angles \(0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\). Cosine is just the reverse!
| Angle (\(\theta\)) | \(\frac{\pi}{6}\) (30°) | \(\frac{\pi}{4}\) (45°) | \(\frac{\pi}{3}\) (60°) |
|---|---|---|---|
| \(\sin \theta\) | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{3}}{2}\) |
| \(\cos \theta\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) | \(\frac{1}{2}\) |
| \(\tan \theta\) | \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\) | \(1\) | \(\sqrt{3}\) |
If you see an equation like \(\tan x = 1\), you must immediately recognise that the principal angle (or reference angle) is \(\frac{\pi}{4}\).
Always work in radians unless the question specifies degrees. Ensure your calculator is in RAD mode when finding arcsin/arccos/arctan for non-exact values!
2. The Core Concept: Finding General Solutions
A General Solution is a formula that accounts for the periodic nature of trigonometric graphs. Since trig functions repeat every \(360^\circ\) (\(2\pi\)) or \(180^\circ\) (\(\pi\)), we use an integer \(n\) to represent any number of full cycles away from the initial solution.
\(n\) must always be defined as an integer (e.g., \(n = 0, \pm 1, \pm 2, ...\)).
Definition: The Principal Value (\(\alpha\))
The Principal Value (\(\alpha\)) is the smallest, non-negative angle (usually found in the first quadrant, \(0 \leq \alpha \leq \frac{\pi}{2}\)) such that \(\sin \alpha = |k|\), \(\cos \alpha = |k|\), or \(\tan \alpha = |k|\). This is the value your calculator returns if you use the inverse trig function on the positive ratio.
2.1. General Solution for Sine: \( \sin x = k \)
The sine wave is symmetrical about the line \(x = \frac{\pi}{2}\). If \(\alpha\) is one solution, the other solution within the cycle \(0\) to \(2\pi\) is \(\pi - \alpha\).
Formula:
$$x = n\pi + (-1)^n \alpha$$
How it works:
- When \(n\) is even (\(0, 2, 4, ...\)), the term \((-1)^n\) is \(+1\), so \(x = \text{even number} \times \pi + \alpha\). This gives you solutions corresponding to the first quadrant angle, repeating every \(2\pi\).
- When \(n\) is odd (\(1, 3, 5, ...\)), the term \((-1)^n\) is \(-1\), so \(x = \text{odd number} \times \pi - \alpha\). This gives you solutions corresponding to the second quadrant angle (\(\pi - \alpha\)), repeating every \(2\pi\).
Imagine the number of cycles (\(n\)). The \((-1)^n\) term is like a switch that flips the direction of the solution within that cycle: start from the left (\(\alpha\)) or reflect off the right (\(\pi - \alpha\)).
2.2. General Solution for Cosine: \( \cos x = k \)
The cosine wave is symmetrical about the y-axis (and subsequent peaks). If \(\alpha\) is one solution, the other solution is \(-\alpha\).
Formula:
$$x = 2n\pi \pm \alpha$$
How it works:
This is simpler! Since the cosine period is \(2\pi\), all solutions are simply the principal solution \(\alpha\) or its negative reflection \(-\alpha\), plus any whole multiple of the period \(2\pi\).
2.3. General Solution for Tangent: \( \tan x = k \)
Tangent is unique because its period is only \(\pi\) (180°).
Formula:
$$x = n\pi + \alpha$$
How it works:
If \(\alpha\) is a solution, simply adding \(\pi\), \(2\pi\), \(3\pi\), etc., will give you the next solutions. You only need one general term for tangent because the solution in the third quadrant is just \(\pi + \alpha\).
🔑 Quick Review: General Solution Formulas (The Trinity)
\(n\) is an integer.
- Sine: \(x = n\pi + (-1)^n \alpha\) (Period \(\pi\), alternating)
- Cosine: \(x = 2n\pi \pm \alpha\) (Period \(2\pi\), two cases)
- Tangent: \(x = n\pi + \alpha\) (Period \(\pi\), simplest formula)
3. Solving Complex Trigonometric Equations
In FP1, the equations often involve a complex argument, like \(\sin(2x)\) or \(\cos(3x - 1)\). The strategy is to treat the entire argument (e.g., \(2x\)) as a single variable first, find its general solutions, and then solve for \(x\).
Step-by-Step Procedure
- Isolate: Get the trigonometric function by itself (e.g., \(\sin(\text{Argument}) = k\)).
- Find Principal Value (\(\alpha\)): Determine the reference angle \(\alpha\) using the positive value of \(k\). Use your knowledge of exact values where possible!
- Apply General Solution: Use the appropriate formula for the function (sin, cos, or tan) on the full argument. (e.g., \(2x = \text{General Solution}\)).
- Solve for \(x\): Algebraically rearrange the equation to isolate \(x\). Remember to divide every term in the General Solution by the coefficient of \(x\).
Example 1: Using Exact Values and Complex Arguments (Cosine)
Solve the equation: \(\cos\left(x + \frac{\pi}{6}\right) = -\frac{1}{\sqrt{2}}\)
Step 1: Isolate & Find \(\alpha\)
The function is already isolated. Since the ratio is negative, we know the solutions are in the 2nd and 3rd quadrants (A-S-T-C rule).
- Reference angle \(\alpha\): We look at \(\cos \alpha = +\frac{1}{\sqrt{2}}\). We know \(\alpha = \frac{\pi}{4}\).
- Actual solutions in \(0 \leq \theta < 2\pi\): Quadrant 2 is \(\pi - \frac{\pi}{4} = \frac{3\pi}{4}\). Quadrant 3 is \(\pi + \frac{\pi}{4} = \frac{5\pi}{4}\).
- (Note: When using the general solution formula for cosine, we only need to use the smallest positive solution, \(\frac{3\pi}{4}\), because the \(\pm\) already accounts for all possibilities relative to the period.)
Step 2: Apply General Solution (to the Argument)
The argument is \(X = x + \frac{\pi}{6}\). We need to find the general solutions for \(\cos X = -\frac{1}{\sqrt{2}}\).
Using the Cosine General Solution: \(X = 2n\pi \pm \beta\), where \(\beta\) is one of the solutions we found (\(\frac{3\pi}{4}\) or \(\frac{5\pi}{4}\)). We usually pick the principal solution for the *equation*, which is \(\frac{3\pi}{4}\).
$$x + \frac{\pi}{6} = 2n\pi \pm \frac{3\pi}{4}$$
Step 3: Solve for \(x\)
We must split this into two cases, using the \(+\) and \(-\) sign.
Case A: Using the + sign
$$x = 2n\pi + \frac{3\pi}{4} - \frac{\pi}{6}$$
Find a common denominator for \(\frac{3\pi}{4} - \frac{\pi}{6}\). LCD is 12.
$$\frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12}$$
Solution A: \(x = 2n\pi + \frac{7\pi}{12}\)
Case B: Using the - sign
$$x = 2n\pi - \frac{3\pi}{4} - \frac{\pi}{6}$$
Find a common denominator for \(-\frac{3\pi}{4} - \frac{\pi}{6}\). LCD is 12.
$$-\frac{9\pi}{12} - \frac{2\pi}{12} = -\frac{11\pi}{12}$$
Solution B: \(x = 2n\pi - \frac{11\pi}{12}\)
(Always remember to state: where \(n\) is an integer.)
Common Mistake to Avoid!
When solving an equation like \(\sin(2x) = k\), a major error is applying the General Solution formula to \(x\) and then multiplying by 2. No!
Wrong way: \(x = n\pi + (-1)^n \alpha\); then \(2x = 2n\pi + 2(-1)^n \alpha\). (Incorrect)
Correct way: \(2x = n\pi + (-1)^n \alpha\); then \(x = \frac{n\pi}{2} + (-1)^n \frac{\alpha}{2}\). (Correct)
Rule: Divide all terms in the general solution expression by the coefficient of \(x\).
Example 2: Handling Non-Exact Values (Sine)
Solve the equation: \(\sin(2x) = 0.3\)
Step 1: Find Principal Value (\(\alpha\))
Since 0.3 is not an exact value, use a calculator (in radians) to find \(\alpha\).
$$\alpha = \arcsin(0.3) \approx 0.3047 \text{ radians}$$
We keep the full argument: \(2x\).
Step 2: Apply General Solution (Sine formula)
$$2x = n\pi + (-1)^n (0.3047...)$$
Step 3: Solve for \(x\)
Divide all terms by 2:
$$x = \frac{n\pi}{2} + (-1)^n \frac{0.3047...}{2}$$
$$x = \frac{n\pi}{2} + (-1)^n (0.1524...)$$
The question will usually specify the required accuracy (e.g., 3 significant figures).
Did You Know? 🤔
These general solutions are incredibly important in fields like signal processing and physics! Whenever you model a wave, such as sound or light, you use periodic functions, and general solutions help engineers predict all possible times a signal will hit a certain peak or zero point.
Key Takeaway: Trigonometry in FP1
To master FP1 trigonometry, focus entirely on the General Solution formulas. Treat the complex argument (e.g., \(2x+A\)) as a single variable, apply the formula, and then carefully isolate \(x\) by dividing all components of the formula by the coefficient of \(x\).