In a right-angled triangle, the three sides are:

• Hypotenuse: The longest side (opposite the right angle)
• Opposite: The side opposite to the angle θ
• Adjacent: The side next to the angle θ (but not the hypotenuse)

For an acute angle θ, the six trigonometric ratios are defined as:

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

The reciprocal ratios are:

$\csc\theta = \frac{1}{\sin\theta} = \frac{\text{hypotenuse}}{\text{opposite}}$

$\sec\theta = \frac{1}{\cos\theta} = \frac{\text{hypotenuse}}{\text{adjacent}}$

$\cot\theta = \frac{1}{\tan\theta} = \frac{\text{adjacent}}{\text{opposite}}$

Pythagorean Identities

From the Pythagorean theorem, we derive the fundamental identities:

$\sin^2\theta + \cos^2\theta = 1$

Dividing by $\sin^2\theta$:

$1 + \cot^2\theta = \csc^2\theta$

Dividing by $\cos^2\theta$:

$1 + \tan^2\theta = \sec^2\theta$

Worked Example 1

Problem: If $\tan\theta = \sqrt{3}$, where θ is an acute angle, find the values of: (a) $\sec\theta$ (b) $\cot\theta$ (c) $\csc\theta$

Solution:

Given $\tan\theta = \frac{\sqrt{3}}{1}$, we can draw a right-angled triangle where:

• Opposite side = $\sqrt{3}$
• Adjacent side = 1

Using the Pythagorean theorem: $\text{Hypotenuse} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Therefore:

• $\sec\theta = \frac{2}{1} = 2$
• $\cot\theta = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
• $\csc\theta = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

These formulae express trigonometric functions of (A ± B) in terms of functions of A and B.

Addition Formulae

$\sin(A + B) = \sin A \cos B + \cos A \sin B$

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Subtraction Formulae

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

$\cos(A - B) = \cos A \cos B + \sin A \sin B$

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Worked Example 2

Problem: Prove that $\cos(x + y)\cos(x - y) = \cos^2x - \sin^2y$

Solution:

Starting with the left-hand side: $\cos(x + y)\cos(x - y)$

Using compound angle formulae: $= (\cos x \cos y - \sin x \sin y)(\cos x \cos y + \sin x \sin y)$

This is in the form (a - b)(a + b) = a² - b²: $= \cos^2x \cos^2y - \sin^2x \sin^2y$

Using $\cos^2y = 1 - \sin^2y$: $= \cos^2x(1 - \sin^2y) - (1 - \cos^2x)\sin^2y$

$= \cos^2x - \cos^2x\sin^2y - \sin^2y + \cos^2x\sin^2y$

$= \cos^2x - \sin^2y$

Proved.

These are special cases of compound angles where A = B:

$\sin 2A = 2\sin A \cos A$

$\cos 2A = \cos^2A - \sin^2A = 2\cos^2A - 1 = 1 - 2\sin^2A$

$\tan 2A = \frac{2\tan A}{1 - \tan^2A}$

Half Angle Formulae

From the double angle formulae, we derive:

$\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2}$

$\cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} = 1 - 2\sin^2\frac{A}{2} = 2\cos^2\frac{A}{2} - 1$

$\tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}}$

Worked Example 3

Problem: Express $\sin 3\alpha$ in terms of $\sin\alpha$

Solution:

$\sin 3\alpha = \sin(2\alpha + \alpha)$ $= \sin 2\alpha \cos\alpha + \cos 2\alpha \sin\alpha$ $= (2\sin\alpha\cos\alpha)\cos\alpha + (1 - 2\sin^2\alpha)\sin\alpha$ $= 2\sin\alpha\cos^2\alpha + \sin\alpha - 2\sin^3\alpha$ $= 2\sin\alpha(1 - \sin^2\alpha) + \sin\alpha - 2\sin^3\alpha$ $= 2\sin\alpha - 2\sin^3\alpha + \sin\alpha - 2\sin^3\alpha$ $= 3\sin\alpha - 4\sin^3\alpha$

Therefore, $\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha$

When θ is very small (measured in radians), we can approximate:

• $\sin\theta \approx \theta$
• $\cos\theta \approx 1 - \frac{1}{2}\theta^2$
• $\tan\theta \approx \theta$

These approximations are useful in physics and engineering for simplifying calculations.

Worked Example 4

Problem: Approximate $\frac{\sin 4\theta - \tan 2\theta}{3\theta}$ when θ is small

Solution:

When θ → 0:

• $\sin 4\theta \approx 4\theta$
• $\tan 2\theta \approx 2\theta$

Therefore: $\frac{\sin 4\theta - \tan 2\theta}{3\theta} \approx \frac{4\theta - 2\theta}{3\theta} = \frac{2}{3}$

These express sums or differences of trigonometric functions as products:

$\sin P + \sin Q = 2\sin\left(\frac{P+Q}{2}\right)\cos\left(\frac{P-Q}{2}\right)$

$\sin P - \sin Q = 2\cos\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right)$

$\cos P + \cos Q = 2\cos\left(\frac{P+Q}{2}\right)\cos\left(\frac{P-Q}{2}\right)$

$\cos P - \cos Q = -2\sin\left(\frac{P+Q}{2}\right)\sin\left(\frac{P-Q}{2}\right)$

Worked Example 5

Problem: Prove that $\sin\theta + \sin 2\theta + \sin 3\theta = \sin 2\theta(2\cos\theta + 1)$

Solution:

$\sin\theta + \sin 2\theta + \sin 3\theta = \sin 2\theta + (\sin\theta + \sin 3\theta)$

Using the factor formula for $\sin\theta + \sin 3\theta$: $= \sin 2\theta + 2\sin\left(\frac{\theta + 3\theta}{2}\right)\cos\left(\frac{\theta - 3\theta}{2}\right)$

$= \sin 2\theta + 2\sin 2\theta \cos(-\theta)$

Since $\cos(-\theta) = \cos\theta$: $= \sin 2\theta + 2\sin 2\theta \cos\theta$

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

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

Proved.

Solving equations of the form $a\cos\theta + b\sin\theta = c$

Method 1: Using t-formulae

Let $t = \tan\frac{\theta}{2}$, then: $\sin\theta = \frac{2t}{1 + t^2}$ $\cos\theta = \frac{1 - t^2}{1 + t^2}$

Method 2: Using R-form

Express $a\cos\theta + b\sin\theta$ in the form $R\cos(\theta - \alpha)$ where: $R = \sqrt{a^2 + b^2}$ $\tan\alpha = \frac{b}{a}$

Worked Example 6

Problem: Solve $5\cos\theta - 2\sin\theta = 2$ for $0° ≤ θ ≤ 360°$

Solution:

Using the R-form method: $R = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}$

$\tan\alpha = \frac{-2}{5} = -0.4$ $\alpha = \tan^{-1}(-0.4) = -21.8°$

So the equation becomes: $\sqrt{29}\cos(\theta + 21.8°) = 2$ $\cos(\theta + 21.8°) = \frac{2}{\sqrt{29}} = 0.371$

Therefore: $\theta + 21.8° = \cos^{-1}(0.371)$

$\theta + 21.8° = 68.3°$ or $291.7°$

$\theta = 46.5°$ or $269.9°$

For trigonometric equations:

• If $\sin\theta = \sin\alpha$, then $\theta = 180°n + (-1)^n\alpha$
• If $\cos\theta = \cos\alpha$, then $\theta = 360°n ± \alpha$
• If $\tan\theta = \tan\alpha$, then $\theta = 180°n + \alpha$

where n is any integer.

Sine Function: $f(x) = \sin x$

• Domain: All real numbers ($x \in \mathbb{R}$)
• Range: $-1 ≤ f(x) ≤ 1$
• Period: $2\pi$ radians

Cosine Function: $f(x) = \cos x$

• Domain: All real numbers ($x \in \mathbb{R}$)
• Range: $-1 ≤ f(x) ≤ 1$
• Period: $2\pi$ radians

Tangent Function: $f(x) = \tan x$

• Domain: All real numbers except $x = \frac{\pi}{2} + n\pi$
• Range: All real numbers ($f(x) \in \mathbb{R}$)
• Period: $\pi$ radians

Worked Example 7

Problem: Find the domain and range of $f(x) = 2\sin x$

Solution:

• Domain: Since sine is defined for all real numbers, domain = $\{x: x \in \mathbb{R}\}$
• Range: Since $-1 ≤ \sin x ≤ 1$, multiplying by 2 gives: $-2 ≤ 2\sin x ≤ 2$
• Range: $\{f(x): -2 ≤ f(x) ≤ 2\}$

Definitions

• $\theta = \sin^{-1}x$ (or $\arcsin x$) if and only if $x = \sin\theta$ with $-\frac{\pi}{2} ≤ θ ≤ \frac{\pi}{2}$
• $\theta = \cos^{-1}x$ (or $\arccos x$) if and only if $x = \cos\theta$ with $0 ≤ θ ≤ \pi$
• $\theta = \tan^{-1}x$ (or $\arctan x$) if and only if $x = \tan\theta$ with $-\frac{\pi}{2} < θ < \frac{\pi}{2}$

Domain and Range of Inverse Functions

Worked Example 8

Problem: Show that $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$

Solution:

Let $x = \sin\theta$, so $\theta = \sin^{-1}x$

Then: $\sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})$ $= \sin^{-1}(2\sin\theta\sqrt{\cos^2\theta})$ $= \sin^{-1}(2\sin\theta\cos\theta)$ $= \sin^{-1}(\sin 2\theta)$ $= 2\theta$ $= 2\sin^{-1}x$

Proved.

The graphs of inverse trigonometric functions are reflections of the original functions across the line $y = x$, with restricted domains.

Key Properties

• $\sin^{-1}x$: Restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, range $[-1, 1]$
• $\cos^{-1}x$: Restricted to $[0, \pi]$, range $[-1, 1]$
• $\tan^{-1}x$: Restricted to $(-\frac{\pi}{2}, \frac{\pi}{2})$, range all real numbers

The graphs have characteristic shapes with asymptotes where applicable.

1. Pythagorean Identities:

   • $\sin^2\theta + \cos^2\theta = 1$
   • $1 + \tan^2\theta = \sec^2\theta$
   • $1 + \cot^2\theta = \csc^2\theta$

2. Compound Angles:

   • $\sin(A ± B) = \sin A \cos B ± \cos A \sin B$
   • $\cos(A ± B) = \cos A \cos B ∓ \sin A \sin B$
   • $\tan(A ± B) = \frac{\tan A ± \tan B}{1 ∓ \