Graphs of the Trigonometric Functions

Picture a point moving anticlockwise around the unit circle at a steady rate, and imagine plotting its height against the angle turned. At which angles is the height greatest, least, and zero? Sketch what you think the graph looks like before reading on.

Plotting each ratio against the angle turns the circle into a curve. Sine and cosine give the same wave, of height $1$ and repeating every $360°$, with cosine a quarter turn ahead. Tangent is different: it breaks where cosine is zero and repeats every $180°$.

To graph $y = \sin\theta$, take the angle along the horizontal axis and the value up the vertical axis. Since $\sin\theta$ is the height of the point on the unit circle, the graph is a record of that height as the point goes round.

Follow one full turn:

at $0°$ the point is at $(1,0)$, so the height is $0$;
rising to $1$ at $90°$, the top of the circle;
back to $0$ at $180°$;
down to $-1$ at $270°$, the bottom;
back to $0$ at $360°$.

Joining these smoothly gives one cycle of a wave. Then the point goes round again, and the whole pattern repeats.

The rate at which the height changes explains the shape. Near $0°$ the point is moving almost straight up, so the height changes quickly and the graph is steep. Near $90°$ it is moving almost sideways, so the height barely changes and the graph flattens at its peak. That is why the graph is a smooth wave rather than a zigzag, and it is the four qualitative shapes from the Rates of Change area appearing in one turn.

The cosine graph is built the same way, plotting the first coordinate instead of the second. Following one turn: it starts at $1$, falls to $0$ at $90°$, to $-1$ at $180°$, back to $0$ at $270°$, and to $1$ at $360°$.

Comparing the two graphs shows they are the same shape. The cosine curve is simply the sine curve slid $90°$ to the left:

$$\cos\theta = \sin(\theta + 90°)$$

The reason is on the circle. The first coordinate of the point at angle $\theta$ is the second coordinate of the point $90°$ further round, because turning a quarter turn exchanges the two axes' roles.

Both graphs share their key features:

period $360°$, since a full turn returns to the start;
amplitude $1$, since the coordinates never exceed $1$ in size;
a maximum of $1$, a minimum of $-1$, and a crossing of the horizontal axis every $180°$.

Where they differ is only where the cycle begins: sine starts at zero and rises, cosine starts at its maximum and falls.

The tangent graph looks nothing like the other two, and the quotient definition explains every difference.

It breaks. $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ is undefined wherever $\cos\theta = 0$, that is at $90°$, $270°$, and every $180°$ from those. At each of those angles the graph has a vertical asymptote: approaching from the left the values grow without bound, and just past it they come up from far below.

It is unbounded. Unlike sine and cosine, tangent takes every real value. Near an asymptote the denominator is tiny while the numerator is close to $1$, so the quotient is enormous.

Its period is $180°$, not $360°$. Adding $180°$ negates both coordinates, and the two minus signs cancel in the quotient, so the value is unchanged. That was proved in Lesson 3.

It always rises. Within each branch the graph increases from far below to far above, crossing the horizontal axis at $0°$, $180°$, $360°$ and so on, where sine is zero.

Reading tangent as the gradient of the radius makes the asymptotes obvious: at $90°$ the radius is vertical, and a vertical line has no gradient.

Negative angles. The graphs continue to the left of zero without any change of rule, since turning clockwise is as legitimate as turning anticlockwise. From Lesson 3, $\cos(-\theta) = \cos\theta$, so the cosine graph is symmetric about the vertical axis: folding the page along that axis matches it onto itself. And $\sin(-\theta) = -\sin\theta$, so the sine graph has rotational symmetry about the origin instead: turning it half a turn about the origin leaves it unchanged.

Counting solutions. The graphs answer "how many angles satisfy this" at a glance. To solve $\sin\theta = 0.5$ for $0° \leq \theta \leq 360°$, draw the horizontal line at height $0.5$ and count crossings: there are two, one on the way up and one on the way down. For $\sin\theta = 1$ there is exactly one, at the peak. For $\sin\theta = 1.4$ there are none, since the line lies above the whole graph.

Over a wider interval there are proportionally more: on $0°$ to $720°$, the equation $\sin\theta = 0.5$ has four solutions, since the pattern repeats once.

Tangent behaves differently again: because it takes every value once per branch, $\tan\theta = c$ has exactly one solution in each $180°$ interval, whatever $c$ is. Lesson 6 uses all of this to solve equations properly.

Five items on the trigonometric graphs. Work each one, then reveal the answer.

1. 1 State the period of $y = \cos\theta$.

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   How far along before it repeats exactly.$360°$

2. 2 State the period of $y = \tan\theta$.

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   Adding $180°$ negates both coordinates, and the signs cancel.$180°$

3. 3 At which angles between $0°$ and $360°$ does $y = \tan\theta$ have asymptotes?

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   Wherever cosine is zero.$90°$ and $270°$

4. 4 State the amplitude of $y = \sin\theta$.

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   Half the distance from minimum to maximum.$1$

5. 5 How many solutions does $\cos\theta = 0$ have for $0° \le \theta \le 360°$?

   Show answer
   Count where the curve crosses the horizontal axis.Two, at $90°$ and $270°$