Trigonometry Made Simple: Sin, Cos, and Tan Explained

Trigonometry is one of the topics students most commonly say they "just don't get." That's almost always because it was introduced too abstractly — as ratios to memorise rather than relationships to understand. Once you see what sin, cos, and tan actually represent, the whole subject becomes more logical.

This guide starts from the basics and builds through to the exam-level content you need for Grade 11 and 12 CAPS and IEB maths.

What Trigonometry Is Actually About

Trigonometry is the study of the relationship between the angles and side lengths of triangles. More specifically: given one angle and one side of a right-angled triangle, you can find any other side. Given two sides, you can find any angle.

That's it. Everything in trig — identities, equations, the sine rule, the cosine rule — is built on that foundation.

Right-Angled Triangles: The Three Ratios

In any right-angled triangle, the sides are named relative to the angle you're working with (call it $\theta$):

• Hypotenuse: the longest side, always opposite the right angle — it doesn't change regardless of which other angle you're looking at
• Opposite: the side directly across from angle $\theta$
• Adjacent: the side next to angle $\theta$ (that isn't the hypotenuse)

The three trigonometric ratios are:

| Ratio | Definition | Memory aid | |---|---|---| | $\sin\theta$ | $\dfrac{\text{opposite}}{\text{hypotenuse}}$ | SOH | | $\cos\theta$ | $\dfrac{\text{adjacent}}{\text{hypotenuse}}$ | CAH | | $\tan\theta$ | $\dfrac{\text{opposite}}{\text{adjacent}}$ | TOA |

Say "SOH-CAH-TOA" until it's automatic. Every right-triangle trig problem starts here.

Critical point: The opposite and adjacent sides change depending on which angle you're working with. The hypotenuse is always the hypotenuse — the other two labels shift.

Worked Example 1: Finding a Side

In a right triangle, the angle at A is $35^\circ$. The hypotenuse is 12 cm. Find the side opposite angle A.

We know the hypotenuse and want the opposite → use sin.

$\sin 35^\circ = \frac{\text{opposite}}{12}$

$\text{opposite} = 12 \times \sin 35^\circ$

$\text{opposite} = 12 \times 0.5736$

$\text{opposite} \approx \mathbf{6.88 \text{ cm}}$

Worked Example 2: Finding an Angle

In a right triangle, the side adjacent to angle $\theta$ is 8 m and the hypotenuse is 10 m. Find $\theta$.

We have adjacent and hypotenuse → use cos.

$\cos\theta = \frac{8}{10} = 0.8$

$\theta = \cos^{-1}(0.8)$

$\theta \approx \mathbf{36.87^\circ}$

The inverse function ($\cos^{-1}$, $\sin^{-1}$, or $\tan^{-1}$) undoes the ratio to give you the angle. On your calculator, this is usually the "2nd" or "shift" function key.

Special Angles — Memorise These

These values come up constantly and are often required in exact form:

| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ | |---|---|---|---| | $0^\circ$ | $0$ | $1$ | $0$ | | $30^\circ$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$ | | $45^\circ$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ | | $60^\circ$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ | | $90^\circ$ | $1$ | $0$ | undefined |

The patterns are worth understanding, not just memorising:

• $\sin\theta$ increases from 0 to 1 as $\theta$ goes $0^\circ$ to $90^\circ$
• $\cos\theta$ decreases from 1 to 0 as $\theta$ goes $0^\circ$ to $90^\circ$
• $\tan 90^\circ$ is undefined because you'd be dividing by $\cos 90^\circ = 0$

The Unit Circle: Beyond Right Triangles

For angles beyond $90^\circ$, we leave the right triangle behind and use the unit circle — a circle of radius 1 centred at the origin of the coordinate plane.

For any angle $\theta$ measured anticlockwise from the positive x-axis:

• $\cos\theta$ = the x-coordinate of the point on the circle
• $\sin\theta$ = the y-coordinate of the point on the circle
• $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{y}{x}$

This extension means $\sin$, $\cos$, and $\tan$ are defined for every angle — not just those between $0^\circ$ and $90^\circ$.

The CAST Diagram

The CAST diagram tells you in which quadrant each ratio is positive:

  II  |  I
 S    |    A
------+------
 T    |    C
 III  |  IV

• Quadrant I ($0^\circ$–$90^\circ$): All ratios are positive
• Quadrant II ($90^\circ$–$180^\circ$): Only Sin is positive
• Quadrant III ($180^\circ$–$270^\circ$): Only Tan is positive
• Quadrant IV ($270^\circ$–$360^\circ$): Only Cos is positive

Memory trick: "All Students Take Calculus" (going anticlockwise from Quadrant I: A, S, T, C).

Using CAST in Problems

If $\sin\theta = 0.6$ and you need all values of $\theta$ between $0^\circ$ and $360^\circ$:

1. Find the reference angle: $\sin^{-1}(0.6) \approx 36.87^\circ$
2. Sin is positive in Quadrants I and II
3. Quadrant I solution: $\theta = 36.87^\circ$
4. Quadrant II solution: $\theta = 180^\circ - 36.87^\circ = 143.13^\circ$

Key Identities

At Grade 11–12 level, you need to know these identities:

Quotient identity:

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

Pythagorean identity:

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

These two come directly from the unit circle definition. They are used to simplify expressions and prove more complex identities.

Co-function identities (to be memorised):

$\sin(90^\circ - \theta) = \cos\theta$

$\cos(90^\circ - \theta) = \sin\theta$

$\sin(180^\circ - \theta) = \sin\theta$

$\cos(180^\circ - \theta) = -\cos\theta$

These appear in simplification and proof questions.

Solving Trigonometric Equations

A trig equation asks for the angle(s) that satisfy a given condition. The method is always the same:

1. Isolate the trig ratio (e.g. $\sin\theta = 0.5$)
2. Find the reference angle using the inverse function
3. Use CAST to identify all quadrants where the ratio has the correct sign
4. Write all solutions in the given domain (usually $0^\circ \leq \theta \leq 360^\circ$, or the general solution for matric)

Worked Example 3: Trig Equation

Solve: $2\cos\theta - 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$

$2\cos\theta = 1$

$\cos\theta = 0.5$

Reference angle: $\cos^{-1}(0.5) = 60^\circ$

Cos is positive in Quadrants I and IV:

• Quadrant I: $\theta = \mathbf{60^\circ}$
• Quadrant IV: $\theta = 360^\circ - 60^\circ = \mathbf{300^\circ}$

Worked Example 4: General Solution

Solve: $\sin\theta = -\dfrac{\sqrt{2}}{2}$, giving the general solution

Reference angle: $\sin^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right) = 45^\circ$

Sin is negative in Quadrants III and IV:

• Quadrant III: $180^\circ + 45^\circ = 225^\circ$
• Quadrant IV: $360^\circ - 45^\circ = 315^\circ$

General solution (adding full rotations):

$\theta = 225^\circ + n \cdot 360^\circ \quad \text{or} \quad \theta = 315^\circ + n \cdot 360^\circ, \quad n \in \mathbb{Z}$

The Sine and Cosine Rules (Non-Right Triangles)

For triangles that are not right-angled, we use the sine rule or cosine rule.

Sine rule:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Use when you have: a side and its opposite angle, plus one more piece of information.

Cosine rule:

$a^2 = b^2 + c^2 - 2bc\cos A$

Use when you have: all three sides (to find an angle), or two sides and the included angle (to find the third side).

Worked Example 5: Sine Rule

In triangle ABC, angle $A = 50^\circ$, angle $B = 70^\circ$, and side $a = 15$ cm. Find side $b$.

Angle $C = 180^\circ - 50^\circ - 70^\circ = 60^\circ$

$\frac{a}{\sin A} = \frac{b}{\sin B}$

$\frac{15}{\sin 50^\circ} = \frac{b}{\sin 70^\circ}$

$b = \frac{15 \times \sin 70^\circ}{\sin 50^\circ}$

$b = \frac{15 \times 0.9397}{0.7660}$

$b \approx \mathbf{18.4 \text{ cm}}$

Common Exam Mistakes

Calculator in radian mode. Virtually every school exam uses degrees. Before each exam, confirm your calculator is in DEG mode (not RAD or GRAD). This single mistake can cost you every trig mark in the paper.

Mixing up opposite and adjacent. Always label the sides relative to the angle you're working with, not the angle you're not working with. A quick diagram takes 10 seconds and eliminates this error.

Forgetting the second solution. Trig equations almost always have two solutions in $[0^\circ, 360^\circ]$. If you find only one, check CAST for the second quadrant.

Rounding the reference angle and then rounding again. Round only at the final answer. If you use $36.87^\circ$ rounded to $36.9^\circ$ partway through, your second-quadrant answer ($180^\circ - 36.9^\circ = 143.1^\circ$) will be slightly off. Keep full precision until the end.

Applying SOH-CAH-TOA to non-right triangles. If there is no right angle in the triangle (or no right angle can be created), you need the sine or cosine rule.

Exam Marks Breakdown

In CAPS Paper 2, trigonometry carries significant weight:

• Grade 11: approximately 50 marks out of 150 across both papers
• Grade 12: trigonometry appears in Paper 2 (identities, equations, 2D/3D problems) — typically 40–60 marks

The most reliable marks come from: correctly identifying the appropriate ratio or rule, correct substitution, and properly handling the general solution or domain restrictions.

Practice Problems

1. In a right triangle, the angle is $42^\circ$ and the adjacent side is 9 cm. Find the hypotenuse.
2. Solve: $\tan\theta = 1$ for $0^\circ \leq \theta \leq 360^\circ$
3. Prove that $\dfrac{\sin^2\theta}{\cos\theta} + \cos\theta = \dfrac{1}{\cos\theta}$
4. In triangle PQR, $PQ = 8$ cm, $PR = 11$ cm, angle $P = 65^\circ$. Find $QR$ (cosine rule).
5. Solve: $2\sin^2\theta - \sin\theta - 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$ (Hint: treat $\sin\theta$ as a variable and factorise.)

Struggling With Trig?

Trigonometry is one of those topics where a couple of targeted sessions can transform your marks. Once the CAST diagram, the identities, and the solution method click into place, the rest follows. If you're finding equations, proofs, or 3D problems difficult, book a session.

I teach Mathematics from Grade 10 through matric for both NSC and IEB, and trig is one of my favourite topics to work through with students.