P2 Chapter 2: Advanced Trigonometry

Solving Trig Puzzles Like a Pro 🧩

Introduction

1. Introduction

sec

\sec(\theta) = \frac{1}{\cos(\theta)}

\sin^2(\theta) + \cos^2(\theta) = 1

Worked example

Time to put the theory into practice! 🚀

2\tan^2(\theta) + \sec(\theta) = 1

1
$\tan^2(\theta)$
$\sec^2(\theta) = 1 + \tan^2(\theta) \implies \tan^2(\theta) = \sec^2(\theta) - 1$
2
$\tan^2(\theta)$
$2(\sec^2(\theta) - 1) + \sec(\theta) = 1$
3
$ax^2 + bx + c = 0$
$2\sec^2(\theta) - 2 + \sec(\theta) = 1 \\ 2\sec^2(\theta) + \sec(\theta) - 3 = 0$
4
$\sec(\theta)$
$(2\sec(\theta) + 3)(\sec(\theta) - 1) = 0 \\ \Rightarrow \sec(\theta) = -\frac{3}{2} \quad \text{or} \quad \sec(\theta) = 1$
5
$\sec^{-1}$
$\cos(\theta) = -\frac{2}{3} \quad \text{or} \quad \cos(\theta) = 1$
6
$\theta$
$\cos(\theta) = 1 \implies \theta = 0^\circ, 360^\circ \\ \cos(\theta) = -\frac{2}{3} \implies \text{Principal Value} = \cos^{-1}(-\frac{2}{3}) \approx 131.8^\circ \\ \text{Second solution (in 3rd quadrant)} = 360^\circ - 131.8^\circ = 228.2^\circ \\ \text{Final Solutions: } \theta = 0^\circ, 131.8^\circ, 228.2^\circ, 360^\circ$

Answer

\cos(\theta) = 1 \implies \theta = 0^\circ, 360^\circ \\ \cos(\theta) = -\frac{2}{3} \implies \text{Principal Value} = \cos^{-1}(-\frac{2}{3}) \approx 131.8^\circ \\ \text{Second solution (in 3rd quadrant)} = 360^\circ - 131.8^\circ = 228.2^\circ \\ \text{Final Solutions: } \theta = 0^\circ, 131.8^\circ, 228.2^\circ, 360^\circ

\sin(A+B)

Why care? This isn't just abstract math. In physics, these formulae are essential for analysing wave interference and superposition. In engineering, they're used in signal processing. And if you're heading into computer science or graphics, understanding how to manipulate angles like this is fundamental. Mastering these sets you up for success in your exams and in many STEM fields at university. 💪

Worked example

Let's Untangle This Trig Puzzle 🧩

x

a\sin\theta

R

Worked example

Let's Solve This Thing! 🚀

4\sin\theta + \cos\theta

1
$R\cos(\theta - \alpha)$
$R\cos(\theta - \alpha) = R(\cos\theta\cos\alpha + \sin\theta\sin\alpha) = (R\cos\alpha)\cos\theta + (R\sin\alpha)\sin\theta$
2
$R$
$R = \sqrt{1^2 + 4^2} = \sqrt{1+16} = \sqrt{17}$
3
$For part (b), we substitute our new harmonic form into the equation. This makes it a much simpler equation to handle.$
$\sqrt{17}\cos(\theta - 75.96^\circ) = 3$
4
$\beta = \theta - 75.96^\circ$
$\beta = \arccos\left(\frac{3}{\sqrt{17}}\right)$
5
$\beta$
$\text{1st Quadrant Solution: } \beta_1 = 43.31^\circ$
6
$\beta$
$\theta_1 = 43.31^\circ + 75.96^\circ = 119.27^\circ \approx 119.3^\circ$

Answer

\theta_1 = 43.31^\circ + 75.96^\circ = 119.27^\circ \approx 119.3^\circ

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