Sketching Sine, Cosine, and Tangent Graphs Revision Notes | GCSE AQA Higher Further Maths

1. Why Trigonometric Graphs Matter

You already know $\sin$, $\cos$, and $\tan$ as ratios in right-angled triangles. But those ratios keep working for any angle — including angles bigger than $90^\circ$ or even negative angles. Plotting them as graphs reveals a beautiful repeating pattern that appears in everything from sound waves to tidal cycles.

At GCSE, you need to be able to sketch all three graphs from memory, label their key features, and use them to read off values or solve equations. The good news: once you know the shape and a handful of coordinates, you can reproduce any of them in under a minute.

2. The Sine Graph: $y = \sin x$

Think of the sine graph as a smooth wave that starts at the origin, rises to a peak, falls back through zero, dips to a trough, and returns — all within $360^\circ$. That full journey is one cycle, and then it repeats forever in both directions.

The graph continues beyond $360^\circ$ — at $450^\circ$ it peaks again at $1$, and at $-90^\circ$ it troughs at $-1$. The wave is perfectly symmetric about the $x$-axis, and the maximum value is always $1$ while the minimum is always $-1$.

The five anchor points — $(0^\circ, 0)$, $(90^\circ, 1)$, $(180^\circ, 0)$, $(270^\circ, -1)$, $(360^\circ, 0)$ — are the skeleton of the sketch. Plot those five, then draw a smooth continuous wave through them. Examiners award marks for the correct shape and correctly labelled intercepts and turning points.

3. The Cosine Graph: $y = \cos x$

The cosine graph has exactly the same wave shape as sine — same period of $360^\circ$, same amplitude of $1$. The key difference is where it starts: cosine begins at its peak of $1$ when $x = 0^\circ$, whereas sine starts at zero.

You can think of the cosine graph as the sine graph shifted $90^\circ$ to the left. That relationship is worth remembering, but for sketching purposes, learn the cosine anchor points directly.

Notice that $y = \cos x$ is symmetric about the $y$-axis — it is an even function, meaning $\cos(-x) = \cos x$. This means the graph looks the same on both sides of the vertical axis. Sine, by contrast, has rotational symmetry about the origin.

4. Comparing Sine and Cosine Side by Side

Placing both graphs on the same axes makes the relationship clear. They are identical waves, offset by $90^\circ$. Both have amplitude $1$ and period $360^\circ$; the only difference is their starting position.

5. The Tangent Graph: $y = \tan x$

The tangent graph behaves very differently from sine and cosine. It has no maximum or minimum value — it shoots off to positive and negative infinity. This happens because $\tan x = \frac{\sin x}{\cos x}$, and whenever $\cos x = 0$ (at $90^\circ$, $270^\circ$, etc.), you are dividing by zero, which is undefined.

At those undefined angles, the graph has vertical asymptotes — dashed vertical lines that the curve approaches but never crosses. The graph repeats every $180^\circ$, so its period is half that of sine and cosine.

Each branch of the tangent graph is a smooth curve that rises from $-\infty$ on the left of each asymptote to $+\infty$ on the right. The curve passes through the origin and increases steeply, and each branch has the same shape. When sketching, draw the asymptotes as dashed lines first — they act as a framework for the curve.

Worked example: Sketching $y = \tan x$ for $-180^\circ \leq x \leq 360^\circ$

Sketch the graph of $y = \tan x$ for $-180^\circ \leq x \leq 360^\circ$, labelling asymptotes and key coordinates.

1. Asymptotes at $x = -90^\circ$, $x = 90^\circ$, $x = 270^\circ$ — These are where $\cos x = 0$, making $\tan x$ undefined. Draw dashed vertical lines at each. [M1 for correct asymptote positions]
2. $\tan x = 0$ at $x = -180^\circ,\ 0^\circ,\ 180^\circ,\ 360^\circ$ — These are where $\sin x = 0$. Mark these points on the $x$-axis. [M1 for correct zero positions]
3. $\tan 45^\circ = 1$, so the point $(45^\circ, 1)$ lies on the curve — This helps position the curve correctly between $0^\circ$ and $90^\circ$. Similarly, $\tan(-45^\circ) = -1$.
4. Between each pair of asymptotes, draw a smooth curve rising from bottom-left to top-right, passing through the zero crossing — Each branch has the same shape. The curve approaches but never touches the asymptotes. [A1 for correct shape with smooth branches]
5. Label asymptotes $x = -90^\circ$, $x = 90^\circ$, $x = 270^\circ$; label zero crossings; label axes — Examiners expect labelled asymptotes and at least the zero crossings to be marked. [A1 for complete labelling]

Answer: A sketch showing three branches of $y = \tan x$, with dashed asymptotes at $x = -90^\circ$, $x = 90^\circ$, $x = 270^\circ$, and zero crossings at $x = -180^\circ$, $0^\circ$, $180^\circ$, $360^\circ$.

6. Sketching All Three: A Worked Method

Now that you know the shape of each graph, here is a reliable method for producing a correct sketch under exam conditions. The key is to plot the anchor points first, then connect them with a smooth curve — never draw a jagged or angular line through trig graphs.

Worked example: Sketching $y = \sin x$ for $0^\circ \leq x \leq 360^\circ$

Sketch the graph of $y = \sin x$ for $0^\circ \leq x \leq 360^\circ$, labelling all key features.

1. Horizontal axis: $x$ from $0^\circ$ to $360^\circ$, marked at $90^\circ$ intervals. Vertical axis: $y$ from $-1$ to $1$. — Correct axis labels and scale earn method marks even if the curve has minor inaccuracies.
2. $(0^\circ, 0)$, $(90^\circ, 1)$, $(180^\circ, 0)$, $(270^\circ, -1)$, $(360^\circ, 0)$ — These five points define the skeleton of the wave. Mark them clearly with dots. [M1 for correct anchor points]
3. Draw a smooth S-shaped curve: rises from $(0^\circ, 0)$ to peak at $(90^\circ, 1)$, falls through $(180^\circ, 0)$ to trough at $(270^\circ, -1)$, rises back to $(360^\circ, 0)$. — The curve must be smooth and continuous — no sharp corners. [A1 for correct wave shape]
4. Label the maximum $(90^\circ, 1)$, minimum $(270^\circ, -1)$, and the $x$-intercepts at $0^\circ$, $180^\circ$, $360^\circ$. — Labelling turning points and intercepts is expected in a sketch question. [A1 for complete labelling]

Answer: A smooth wave from $(0^\circ, 0)$ peaking at $(90^\circ, 1)$, crossing zero at $(180^\circ, 0)$, troughing at $(270^\circ, -1)$, and returning to $(360^\circ, 0)$.

7. Using Trig Graphs to Find Values and Solve Equations

Trig graphs are not just for sketching — they let you find all solutions to equations like $\sin x = 0.5$ in a given interval. The graph shows you visually how many times the horizontal line $y = 0.5$ crosses the curve, and where.

For example, $\sin x = 0.5$ has solutions at $x = 30^\circ$ and $x = 150^\circ$ in the interval $0^\circ \leq x \leq 360^\circ$. The graph makes it clear why: the line $y = 0.5$ cuts the sine wave at two points in that range. Without the graph, students often miss the second solution entirely.

Worked example: Using the sine graph to find all solutions in an interval

Given that $\sin 30^\circ = 0.5$, use the graph of $y = \sin x$ to find all values of $x$ in the interval $0^\circ \leq x \leq 360^\circ$ for which $\sin x = 0.5$.

1. $x = 30^\circ$ (given) — This is the acute angle from the calculator or given information. [B1]
2. The sine graph is symmetric about $x = 90^\circ$ in the first half-cycle. Second solution: $x = 180^\circ - 30^\circ = 150^\circ$ — The sine wave rises to its peak at $90^\circ$ and is symmetric either side. So if $\sin 30^\circ = 0.5$, then $\sin 150^\circ = 0.5$ as well. [M1 for correct use of symmetry]
3. The next time $\sin x = 0.5$ would be at $30^\circ + 360^\circ = 390^\circ$, which is outside the interval. — Always check whether additional cycles fall within the required interval. [A1 for both solutions stated]
4. $x = 30^\circ$ and $x = 150^\circ$ — Both values satisfy $\sin x = 0.5$ in $0^\circ \leq x \leq 360^\circ$.

Answer: $x = 30^\circ$ and $x = 150^\circ$

The same principle applies to cosine and tangent. For $\cos x = 0.5$, the graph shows solutions at $x = 60^\circ$ and $x = 300^\circ$ in $0^\circ \leq x \leq 360^\circ$. For tangent, because the period is $180^\circ$, there is exactly one solution per $180^\circ$ interval.

8. Practice: Sketch and Retrieve

The best way to lock in these graphs is to reproduce them from scratch. Cover the page and try each one before checking. Focus on getting the shape right and the five anchor points in the correct positions.

9. Exam Technique and Common Mistakes

Sketch questions are worth marks for shape, scale, and labelling — not just for getting the curve roughly right. Examiners follow a mark scheme that checks specific features, so knowing what they look for is as important as knowing the graph itself.

The most common error is confusing the sine and cosine graphs. Remember: sine starts at zero, cosine starts at one. If you draw a wave that starts at $(0^\circ, 1)$ and label it $y = \sin x$, you will lose marks even if the shape is perfect.