P1 Trigonometry: Graphs, Identities, and Equations

Trig: More Than Just SOH CAH TOA 😎

Introduction

1. Introduction

Alright, let's dive into Trigonometry! You've probably met sin, cos, and tan before, but at A-Level, we take it to a whole new level. We're going to start by visualizing these functions as graphs, understanding their wave-like patterns and how to transform them. Then, we'll ditch the calculator and master the exact trigonometric values for key angles - super useful for cleaner, more professional solutions. After that, we'll become mathematical detectives, proving trigonometric identities which are the fundamental rules that govern these functions. Finally, we'll put it all together to solve complex trigonometric equations, a skill that's a massive part of your P1 exam. It might seem like a lot, but we'll break it down step-by-step. Let's get started!

\frac{\sqrt{3}}{2}

\sin(45^{\circ}) = \frac{O}{H} = \frac{1}{\sqrt{2}}

\sqrt{2^2 - 1^2} = \sqrt{3}

Worked example

Worked Example: Let's Solve This Thing! 🚀

\sin(60^{\circ})\cos(30^{\circ}) + \tan^2(45^{\circ})

\theta

a^2 + b^2 = c^2

Worked example

Time to Be a Mathematical Detective 🕵️‍♀️

\frac{1+\cos\theta}{\sin\theta} + \frac{\sin\theta}{1+\cos\theta} \equiv \frac{2}{\sin\theta}

\sin(\theta) = 0.5

\sin(\theta) = 0.5

Worked example

Let's Solve This Thing! 💪

2\cos(2x + 40^\circ) = \sqrt{2}

1
$(2x + 40^\circ)$
$\cos(2x + 40^\circ) = \frac{\sqrt{2}}{2} \\ \text{Original interval: } 0^\circ \le x \le 360^\circ \\ \text{Adjusted interval: } 0^\circ \le 2x \le 720^\circ \\ 40^\circ \le 2x + 40^\circ \le 760^\circ$
2
$A = 2x + 40^\circ$
$A = \arccos\left(\frac{\sqrt{2}}{2}\right) = 45^\circ$
3
$Using the CAST diagram, we know that cosine is positive in Quadrant 1 and Quadrant 4. So, we find the second base solution.$
$\text{Q1 solution: } A = 45^\circ \\ \text{Q4 solution: } A = 360^\circ - 45^\circ = 315^\circ$
4
$A$
$A_1 = 45^\circ \\ A_2 = 315^\circ \\ A_3 = 45^\circ + 360^\circ = 405^\circ \\ A_4 = 315^\circ + 360^\circ = 675^\circ \\ \text{All four values are within the interval } [40^\circ, 760^\circ].$
5
$A = 2x + 40^\circ$
$2x + 40^\circ = 45^\circ \implies 2x = 5^\circ \implies x = 2.5^\circ \\ 2x + 40^\circ = 315^\circ \implies 2x = 275^\circ \implies x = 137.5^\circ \\ 2x + 40^\circ = 405^\circ \implies 2x = 365^\circ \implies x = 182.5^\circ \\ 2x + 40^\circ = 675^\circ \implies 2x = 635^\circ \implies x = 317.5^\circ$
6
$State the final solutions clearly. All four values are valid.$
$x = 2.5^\circ, 137.5^\circ, 182.5^\circ, 317.5^\circ$

Answer

x = 2.5^\circ, 137.5^\circ, 182.5^\circ, 317.5^\circ

y = \sin(x)

y = \tan(x)

Worked example

Let's Give This Graph a Glow-Up ✨

y = 2 \cos(x) - 1

1
$y = \cos(x)$
$y = 2\cos(x) - 1$
2
$|a|$
$\text{Amplitude} = |a| = |2| = 2$
3
$y=0$
$\text{New Max} = (\text{Old Max} \times a) + d = (1 \times 2) - 1 = 1 <br> \text{New Min} = (\text{Old Min} \times a) + d = (-1 \times 2) - 1 = -3$
4
$y=\cos(x)$
$\text{Maximum points: } (0^\circ, 1) \text{ and } (360^\circ, 1) <br> \text{Minimum point: } (180^\circ, -3)$
5
$y=-1$
$\text{Final Answer: Amplitude} = 2. \text{ Max points are } (0^\circ, 1), (360^\circ, 1). \text{ Min point is } (180^\circ, -3).$

Answer

\text{Final Answer: Amplitude} = 2. \text{ Max points are } (0^\circ, 1), (360^\circ, 1). \text{ Min point is } (180^\circ, -3).

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