Trigonometry: Quadrants, Complementary Angles and Sum-Difference Formulas

How the ratios of large angles are defined, which ratio is positive in which quadrant, and the values at $0^{\circ }$, $90^{\circ }$, $180^{\circ }$, $270^{\circ }$.

The idea

Take a circle with centre at the origin $O$ and radius $1$ (the unit circle). Measure the angle $\theta$ anticlockwise from the positive $x$-axis; the tip of the radius is the point $P=(x,y)$. Definition: $\cos \theta =x$, $\sin \theta =y$, $\tan \theta =\frac{y}{x}$. This is the old definition in disguise: for an acute angle drop a perpendicular from $P$ to the $x$-axis and you get base $=x$, perpendicular $=y$, hypotenuse $=1$, so $\frac{\text{perp}}{\text{hyp}}=y$ and $\frac{\text{base}}{\text{hyp}}=x$.

Now at an angle beyond $90^{\circ }$ the numbers $x$ and $y$ still exist; they may simply be negative. That is the whole secret of the signs: $x$ is negative on the left, $y$ is negative below the axis.

Letters in the table: $x$ = horizontal distance of the point (right is $+$), $y$ = vertical distance (up is $+$). $\sec$, $\mathrm{cosec}$, $\cot$ keep the sign of their reciprocals ($\cos$, $\sin$, $\tan$).

Angles that fall on an axis

$\tan 90^{\circ }=\frac{1}{0}$, and division by zero is impossible, so $\tan 90^{\circ }$ and $\tan 270^{\circ }$ are not defined (they have no value; saying "infinity" is not right either). The table can be read straight from the figure: at $90^{\circ }$ the point is at $(0,1)$.

Steps to find a sign

1. Bring the angle into $0^{\circ }$–$360^{\circ }$ (adding or subtracting full turns of $360^{\circ }$).
2. See which quadrant it is in (the borders are $90^{\circ }$, $180^{\circ }$, $270^{\circ }$).
3. Read the sign of that ratio from ASTC: if its letter is there, $+$, else $-$.
4. For a product of several factors count the negative signs: even $\Rightarrow +$, odd $\Rightarrow -$.

Practice

Values for $180^{\circ }-\theta$, $180^{\circ }+\theta$, $360^{\circ }-\theta$ and the reference angle of any angle.

The idea

Take the point $P=(\cos \theta ,\sin \theta )$ for an acute angle $\theta$. Take three mirror images of it: in the $y$-axis (to the left), through the origin (opposite) and in the $x$-axis (below). They sit exactly at the angles $180^{\circ }-\theta$, $180^{\circ }+\theta$ and $360^{\circ }-\theta$, and each point makes the same angle $\theta$ with the $x$-axis. A mirror image changes only the direction of $x$ or $y$, not the distance, so the size of $\cos$ and $\sin$ stays the same and only the sign can change.

Here: $\theta$ is any acute angle. The $\tan$ column is $\sin ÷\cos$ (at $180^{\circ }-\theta$: $\frac{+\sin }{-\cos }=-\tan$).

Steps

1. Bring the angle into $0^{\circ }$–$360^{\circ }$ and find its quadrant.
2. Write the sign of the given ratio in that quadrant from ASTC.
3. Find the reference angle and write the size from the standard table.
4. Join the sign and the size.

Practice

Ratios of $90^{\circ }\pm \theta$ and $270^{\circ }\pm \theta$: a $90^{\circ }$ turn swaps $\sin$ and $\cos$.

The idea

Turn the point $P=(x,y)$ through $90^{\circ }$ about the origin (anticlockwise). Its new position is $(-y,x)$: the old height has become a distance to the left and the old base has become the height. Watch the small triangle at $P$ being turned in the figure — it is the same triangle, only the roles of its two legs have changed.

Three turns, that is $270^{\circ }$, take the point to $(y,-x)$, so $\cos (270^{\circ }+\theta )=\sin \theta$ and $\sin (270^{\circ }+\theta )=-\cos \theta$. For $90^{\circ }-\theta$ and $270^{\circ }-\theta$ put $-\theta$ in these and use the negative-angle rules (next method).

Here: $\theta$ is an acute angle. For $\sec$, $\mathrm{cosec}$ and $\cot$ use the rule of their reciprocals ($\sec \leftrightarrow \mathrm{cosec}$, $\tan \leftrightarrow \cot$).

Six steps

1. Write the angle as "a multiple of $90^{\circ }$ $\pm \theta$".
2. Is the multiple even ($180$, $360$) or odd ($90$, $270$)? Even: same name; odd: swap the name.
3. Treat $\theta$ as acute and find the quadrant of the whole angle.
4. Read the sign of the given (original) ratio in that quadrant from ASTC.
5. Combine the name and the sign to write the answer.
6. Check: put $\theta =30^{\circ }$ and compare the values of both sides.

Practice

Angles measured clockwise, removing full turns, and the periodicity of $\sin$, $\cos$, $\tan$.

The idea

An angle measured anticlockwise is positive; measured clockwise it is negative. The point at $-\theta$ is the mirror image of the point at $\theta$ in the $x$-axis: the same $x$, the opposite $y$. So $\cos (-\theta )=\cos \theta$ and $\sin (-\theta )=-\sin \theta$. $\tan$ is their quotient, so $\tan (-\theta )=-\tan \theta$.

Now large angles. After a full turn ($360^{\circ }$) the point is exactly where it started, so $x$ and $y$ are the same. Every ratio repeats after $360^{\circ }$. $\tan$ and $\cot$ even repeat every $180^{\circ }$ (because $\tan (180^{\circ }+\theta )=+\tan \theta$).

Practice

The graphs of $\sin$ and $\cos$ traced by the moving point, their symmetries, and comparing angles and counting solutions from the graph.

The idea

A graph is a picture that shows the value for every angle at once. Put the angle $\theta$ along the $x$-axis and $\sin \theta$ as the height. In the figure the pink point turns on the circle at the left; its height (the saffron segment) becomes the height of the graph on the right. As the point turns, the graph is drawn as a wave.

Link: the graph of $\cos \theta$ is the graph of $\sin \theta$ shifted $90^{\circ }$ to the left, because $\cos \theta =\sin (90^{\circ }+\theta )$. That is why the two waves have the same shape, one ahead of the other.

Practice