Sketch the graphs of $y = \sin x$, $y = \cos x$ and $y = \tan x$ in degrees over one or more periods, identify their amplitude (where it exists) and period and the key maximum, minimum, zero and intercept points, and solve trigonometric equations of the form $\sin x = k$, $\cos x = k$ and $\tan x = k$ for all solutions in a given domain using the graph together with the related acute angle and the ASTC rule

What this dot point is asking

The previous pages pinned down what $\sin$, $\cos$ and $\tan$ mean for any angle, using the unit circle and the ASTC sign rule. This dot point steps back and looks at the whole picture: as the angle $x$ runs from $0\degree$ upwards, the three ratios trace out curves, and those curves are the most important shapes in all of senior mathematics. Sine and cosine are waves, and the exam expects you to be able to sketch them in degrees, name their amplitude and period, mark their key points (the peaks, troughs and crossings), and then use that picture to solve equations such as $\sin x = k$, $2\cos x = 1$ or $\tan x = k$ for every solution in a stated range like $0\degree \le x \le 360\degree$.

The skill worth the marks is not drawing a pretty curve. It is two-fold. First, knowing the standard shapes cold, so you can read off how high a wave goes and how often it repeats. Second, a reliable method for all the solutions, not just the one a calculator hands back: the calculator gives a single angle, but a trigonometric equation usually has several answers in a full revolution, and the graph plus the related acute angle and ASTC is how you find them all without missing any. Everything here is in degrees; radians and the calculus of these curves come later.

The answer

The sine and cosine graphs

Imagine a point travelling anticlockwise around the unit circle. Its height above the centre is $\sin x$ and its horizontal position is $\cos x$, where $x$ is the angle turned. Plotting that height against the angle traces $y = \sin x$; plotting the horizontal position traces $y = \cos x$. Because the point returns to where it started after one full turn of $360\degree$, both curves repeat every $360\degree$: they are periodic.

Read the sine curve across one revolution. It starts at $0$ (the point is level with the centre), climbs to its highest value $+1$ at $90\degree$ (the point is at the top), comes back to $0$ at $180\degree$, drops to its lowest value $-1$ at $270\degree$ (the point is at the bottom), and returns to $0$ at $360\degree$. The cosine curve has the same wave shape but starts at its peak: $+1$ at $0\degree$, down to $0$ at $90\degree$, down to $-1$ at $180\degree$, back to $0$ at $270\degree$, and up to $+1$ at $360\degree$. In fact cosine is just the sine wave shifted $90\degree$ to the left, which is why they look identical apart from their starting point.

The dots mark the points you must get right when you sketch: for sine, $(0\degree, 0)$, $(90\degree, 1)$, $(180\degree, 0)$, $(270\degree, -1)$, $(360\degree, 0)$; for cosine, $(0\degree, 1)$, $(90\degree, 0)$, $(180\degree, -1)$, $(270\degree, 0)$, $(360\degree, 1)$. Plot those five points and join them with a smooth wave and your sketch will earn full marks.

Amplitude and period

Two numbers describe a wave. The amplitude is how far it rises above (and falls below) its centre line, the greatest height of the wave. For $y = \sin x$ and $y = \cos x$ the wave reaches $+1$ and $-1$ about the $x$-axis, so the amplitude is $1$. The period is the horizontal length of one complete wave before the pattern repeats; for both basic curves that is $360\degree$, one full revolution.

You can read these straight from an equation in the form $y = a\sin x$ or $y = a\cos x$ without drawing anything. The number $a$ in front multiplies every height, so the amplitude is $|a|$ (its size, always taken positive). For example $y = 3\cos x$ has amplitude $3$, peaking at $+3$ and dipping to $-3$, while its period is unchanged at $360\degree$. Reading the period off a graph is the reverse: measure from one peak to the next peak (or any feature to the next identical feature) and that distance is the period.

The tangent graph

Tangent behaves quite differently because $\tan x = \dfrac{\sin x}{\cos x}$. Wherever $\cos x = 0$, at $90\degree$ and $270\degree$ in the first revolution, the tangent is undefined and the graph shoots up towards a vertical line it never touches, a vertical asymptote. Between the asymptotes the curve rises continuously from far below to far above, passing through $0$ wherever $\sin x = 0$, that is at $0\degree$, $180\degree$ and $360\degree$. Because this whole pattern completes in half a revolution, the period of tangent is $180\degree$, not $360\degree$. Since the curve has no maximum or minimum height, talking about its amplitude makes no sense.

Solving a trigonometric equation for all solutions

Now use the graphs. To solve an equation like $\cos x = k$ over a range is to ask: at which angles does the curve reach the height $k$? Drawing the horizontal line $y = k$ across the wave, every place the line crosses the curve is a solution, and the $x$-coordinate of each crossing is an answer. The line shows you at a glance how many solutions there are: a wave at height between $-1$ and $1$ is cut twice per revolution, so most equations have two answers in $0\degree$ to $360\degree$.

You do not have to read the crossings off a hand-drawn graph, though. The exact-value method combines the related acute angle with the ASTC sign rule from the previous page, and it never misses a solution:

1. Rearrange to get a single ratio alone: $\sin x = k$, $\cos x = k$ or $\tan x = k$.
2. Find the related acute angle from the size of $k$ (ignore any minus sign): use a special angle if $k$ is exact, otherwise the calculator. Never enter a negative number here.
3. Choose the quadrants from the sign of $k$ using ASTC: a positive sine sits in quadrants 1 and 2, a positive cosine in 1 and 4, a positive tangent in 1 and 3; a negative ratio sits in the other two quadrants.
4. Build each in-range angle from the related angle $\alpha$: quadrant 1 gives $\alpha$, quadrant 2 gives $180\degree - \alpha$, quadrant 3 gives $180\degree + \alpha$, quadrant 4 gives $360\degree - \alpha$. Keep only those inside the stated domain, and if the domain runs past $360\degree$ add $360\degree$ to each first-revolution answer to reach the rest.

Take $2\cos x = 1$ over $0\degree \le x \le 360\degree$. Rearranged, $\cos x = \tfrac12$; the related acute angle is $60\degree$ (since $\cos 60\degree = \tfrac12$); the cosine is positive, so $x$ is in quadrants 1 and 4; that gives $60\degree$ and $360\degree - 60\degree = 300\degree$. The horizontal line $y = \tfrac12$ confirms it, cutting the cosine wave at exactly those two angles.

How many solutions, and the boundary cases

Counting solutions is itself an exam skill, and the graph settles it instantly. Slide the line $y = k$ up and down the sine or cosine wave:

• If $-1 < k < 1$ the line crosses twice in each $360\degree$, so the equation has two solutions per revolution.
• If $k = 1$ or $k = -1$ the line just touches the very top or bottom of the wave, meeting it once per revolution (a single solution, a boundary angle such as $\cos x = -1$ at $x = 180\degree$).
• If $k > 1$ or $k < -1$ the line lies entirely above or below the wave and never meets it, so there are no solutions, because $\sin$ and $\cos$ can never leave the range $-1$ to $1$.

This is why an equation like $\sin x = 1.4$ has no solution at all, and why the size of the domain matters: doubling the range to $720\degree$ roughly doubles the number of crossings. The diagram below shows the negative case, $\sin x = -\tfrac12$: the line sits below the axis, so it only meets the parts of the sine wave that dip below zero, giving the two solutions $210\degree$ and $330\degree$ (related angle $30\degree$, sine negative so quadrants 3 and 4).

How exam questions ask about trigonometric graphs and equations

The wording tells you what the markers want:

• "Sketch the graph of $y = \sin x$ for $0\degree \le x \le 360\degree$" wants the wave with the five key points marked and the axes scaled in degrees. Plot $(0,0)$, $(90,1)$, $(180,0)$, $(270,-1)$, $(360,0)$ for sine (or the cosine set) and join smoothly.
• "State the amplitude and period" wants the two numbers only: amplitude $|a|$ from the coefficient, period $360\degree$ for $\sin$ or $\cos$ (or read peak-to-peak off a given graph).
• "Solve ... for $0\degree \le x \le 360\degree$" wants every angle in that range, not just the calculator's first answer. Show the related acute angle and the quadrants, and list all in-range solutions.
• "Find the exact value(s)" signals a special angle ($30\degree$, $45\degree$, $60\degree$ and their relatives), so keep surds like $\dfrac{\sqrt3}{2}$ rather than decimals.
• "Correct to the nearest degree" signals a calculator value; keep full precision until the final rounding.
• "How many solutions" is answered by the line-cutting-the-curve picture: count the crossings of $y = k$ on the wave over the given domain.
• "For $0\degree \le x \le 720\degree$" (or any range past one revolution) wants you to find the first-revolution answers and then add $360\degree$ to each to reach the rest of the domain.
• A real-world model like $d = a + b\cos(nt)\degree$ (tides, temperature, a Ferris wheel) asks for greatest/least values from $a \pm |b|$ and the period from when the inside angle reaches $360\degree$.