IB Mathematics AI AHL Trigonometric equations MAI Study Notes- New Syllabus

IB Mathematics AI AHL Trigonometric equations MAI Study Notes

LEARNING OBJECTIVE

The definitions of cosθ and sinθ in terms of the unit circle.

Key Concepts:

The Pythagorean identity
Definition of tanθ

SIN $\theta$, COS $\theta$

Consider the unit circle (radius $r = 1$) on the Cartesian plane. Let $P(x, y)$ be a point on the circle, $OP = r = 1$, and $\theta$ be the angle between $OP$ and the positive $x$-axis.

Then:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{1} = y$
$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{1} = x$

Thus:

$\sin \theta = y \text{-coordinate of } \theta$
$\cos \theta = x \text{-coordinate of } \theta$

Graphical Representation:

The $y$-axis (showing $\sin \theta$) can be visualized to the left of the circle:

This explains why supplementary angles have equal sines:

$\sin 30^\circ = \sin 150^\circ = 0.5$

Similarly, the $x$-axis (showing $\cos \theta$) can be visualized below the circle:

This explains why opposite angles have equal cosines:

$\cos 60^\circ = \cos 300^\circ = 0.5$

NOTE:

Any point on the circle has infinitely many angle values, all with the same sine and cosine. For example:

$\sin 30^\circ = \sin 390^\circ = 0.5$
$\cos 30^\circ = \cos 390^\circ = \frac{\sqrt{3}}{2}$

PROPERTIES OF SIN $\theta$ AND COS $\theta$

Range:

$-1 \leq \sin \theta \leq 1$
$-1 \leq \cos \theta \leq 1$

Sign in Quadrants:

$\sin \theta$: Positive in $1^\text{st}$ and $2^\text{nd}$ quadrants; negative in $3^\text{rd}$ and $4^\text{th}$ quadrants.
$\cos \theta$: Positive in $1^\text{st}$ and $4^\text{th}$ quadrants; negative in $2^\text{nd}$ and $3^\text{rd}$ quadrants.

Pythagorean Identity:

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

TAN $\theta$:

The trigonometric function $\tan \theta$ is defined as:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

It can take any real value:

$-\infty < \tan \theta < +\infty$

Positive in $1^\text{st}$ and $3^\text{rd}$ quadrants.
Negative in $2^\text{nd}$ and $4^\text{th}$ quadrants.

TRIGONOMETRIC EQUATIONS

A trigonometric equation is an equation that involves one or more trigonometric functions (like sine, cosine, tangent) of a variable, usually an angle. Solving a trigonometric equation means finding all angle values (within a given interval or in general form) that satisfy the equation.

Because trigonometric functions are periodic, solutions often repeat in cycles and are expressed using a general form involving an integer parameter, usually denoted as $k \in \mathbb{Z}$.

Example:

$
\sin x = \frac{1}{2}
$

From known values:

$
\sin x = \frac{1}{2} \Rightarrow x = 30^\circ \quad \text{or} \quad x = 150^\circ
$

Since the sine function has a period of $360^\circ$, the full set of solutions is:

$
x = 30^\circ + 360^\circ k \quad \text{or} \quad x = 150^\circ + 360^\circ k, \quad k \in \mathbb{Z}
$

In radians:

$
x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}
$

Example

Solve

$\sin x = \frac{1}{2}$

for $0 \leq x \leq 2\pi$.

▶️Answer/Explanation

Solution:

Using a GDC, the solutions are:

$x = \frac{\pi}{6} \approx 0.524 \quad \text{and} \quad x = \frac{5\pi}{6} \approx 2.618$

A trigonometric function is a function that relates the angles of a right triangle to the ratios of its sides. These functions are also used to describe periodic phenomena such as sound waves, light waves, and circular motion.

$ f(x) = \sin x $

The graph of $ f(x) = \sin x $ is constructed as follows:

Properties:

Domain: $ x \in \mathbb{R} $
Range: $ y \in [-1, 1] $
Central Axis: $ y = 0 $
Amplitude: $ 1 $
Period: $ T = 2\pi $

The graph oscillates between $-1$ and $1$, completing one full cycle every $2\pi$ units.

$ f(x) = \cos x $

The graph of $ f(x) = \cos x $ is constructed as follows:

Properties:

Domain: $ x \in \mathbb{R} $
Range: $ y \in [-1, 1] $
Central Axis: $ y = 0 $
Amplitude: $ 1 $
Period: $ T = 2\pi $

Graph:

The graph oscillates between $-1$ and $1$, completing one full cycle every $2\pi$ units.

VERTICAL TRANSFORMATIONS

Vertical Translation

For $ f(x) = \sin x + C $:
The graph shifts vertically by $C$ units.
Central Axis: $ y = C $
Range: $ y \in [C – 1, C + 1] $

Vertical Stretch

For $ f(x) = A \sin x $:
The amplitude becomes $ |A| $.
Range: $ y \in [-|A|, |A|] $

Horizontal Stretch

For $ f(x) = \sin(Bx) $:
The period becomes $ T = \frac{2\pi}{|B|} $.

General Form

For

$ f(x) = A \sin(Bx) + C $:

Amplitude: $ |A| $
Period: $ T = \frac{2\pi}{|B|} $
Central Axis: $ y = C $
Range: $ y \in [C – |A|, C + |A|] $

Example

Express the equation as a trigonometric function:

Central Axis: $ y = 5 $
Amplitude: 1.5
Period: $ \pi $
Type: $ -\sin x $

▶️Answer/Explanation

Solution:

$f(x) = -1.5 \sin(2x) + 5$
$f(x) = 1.5 \cos\left(2x – \frac{\pi}{2}\right) + 5$

HORIZONTAL TRANSFORMATIONS

For

$ f(x) = \sin(x – D) $:

The graph shifts horizontally by $D$ units to the right.

For

$ f(x) = A \sin[B(x – D)] + C $:

Horizontal Shift: $ D $
Amplitude: $ |A| $
Period: $ T = \frac{2\pi}{|B|} $
Central Axis: $ y = C $

Example

Given $ f(x) = 5 \sin\left[2\left(x – \frac{\pi}{4}\right)\right] + 7 $

Find

Shift:
Amplitude:
Period:
Central Axis:

▶️Answer/Explanation

Solution:

Shift: $ \frac{\pi}{4} $ units right
Amplitude: 5
Period: $ \pi $
Central Axis: $ y = 7 $

THE AMBIGUOUS CASE

If we are given two sides and a non-included angle, we may have as a solution:

Two triangles
One triangle
No triangle at all

This is because the sine rule provides two values for an unknown angle. For example, if we find $ \sin C = 0.5 $, then:

$ C = 30^\circ \quad (\text{this is } \sin^{-1} C) $
or
$ C’ = 180^\circ – 30^\circ = 150^\circ. $

These two values may result in different solutions.

Example (given two sides and a non-included angle)

Find Solution of triangle.

▶️Answer/Explanation

Solution:

We use the sine rule:
$ \frac{4}{\sin 30^\circ} = \frac{5}{\sin C} \Rightarrow \sin C = 0.625 $
Hence:
$ C = 38.7^\circ \, (\text{by GDC}) $
or
$ C’ = 180^\circ – 38.7^\circ = 141.3^\circ. $

CASE (1): If $ C = 38.7^\circ $, then:
$ A = 180^\circ – 30^\circ – 38.7^\circ, \text{ thus } A = 111.3^\circ $
and then:
$ BC^2 = 5^2 + 4^2 – 2(5)(4)\cos 111.3^\circ \Rightarrow BC = 7.45 $

CASE (2): If $ C’ = 141.3^\circ $, then:
$ A’ = 180^\circ – 30^\circ – 141.3^\circ, \text{ thus } A’ = 8.7^\circ $
and then:
$ BC’^2 = 5^2 + 4^2 – 2(5)(4)\cos 8.7^\circ \Rightarrow BC’ = 1.21 $

We may, sometimes, obtain no solution at all.

Example(given 2 sides and a non-included angle)

Find Solution of triangle.

▶️Answer/Explanation

Solution:

We use the sine rule:
$ \frac{1}{\sin 30^\circ} = \frac{5}{\sin C} \Rightarrow \sin C = 2.5 $
which is impossible!
Hence, there is no such triangle!

◆ JUSTIFICATION OF THE AMBIGUOUS CASE
In Example , we were given $ B=30^\circ $, $ AB=5 $, $ AC=4 $, and we found two solutions for $ C $, and thus two possible triangles: $ ABC $ and $ ABC’ $.