Trigonometric functions, graphs and identities

From triangles to the unit circle

Right triangles only define the ratios for angles between 0° and 90°. To go further, draw the unit circle: centre at the origin, radius 1. Measure an angle $x$ from the positive $x$-axis, anticlockwise for positive angles. The radius at angle $x$ ends at a point $P$, and by definition

$$\cos x = \text{the } x\text{-coordinate of } P \qquad \sin x = \text{the } y\text{-coordinate of } P \qquad \tan x = \frac{\sin x}{\cos x}$$

For acute angles this agrees with the triangle definitions, because the hypotenuse is the radius, 1. Since $P$ is on a circle of radius 1, Pythagoras gives the fundamental identity:

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

Here $\sin^2 x$ means $(\sin x)^2$. This identity turns any known sine into a cosine up to its sign, and the sign comes from the quadrant.

Signs by quadrant

Quadrant	Angles	sin	cos	tan
I	$0$ to $\frac{\pi}{2}$	+	+	+
II	$\frac{\pi}{2}$ to $\pi$	+	−	−
III	$\pi$ to $\frac{3\pi}{2}$	−	−	+
IV	$\frac{3\pi}{2}$ to $2\pi$	−	+	−

To evaluate a ratio of an angle outside the first quadrant, find its reference angle, the acute angle between the radius and the $x$-axis, take the value from the table of special angles, then attach the sign of the quadrant. For $\frac{5\pi}{6}$ (150°) the reference angle is $\frac{\pi}{6}$, so $\sin\frac{5\pi}{6} = \frac{1}{2}$ and $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$. Behind this sit the symmetry rules $\sin(\pi - x) = \sin x$, $\cos(\pi - x) = -\cos x$, $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$.

The graphs, in words

• Sine starts at 0 when $x = 0$, rises to 1 at $\frac{\pi}{2}$, returns to 0 at $\pi$, falls to $-1$ at $\frac{3\pi}{2}$ and is back at 0 at $2\pi$: a smooth wave, repeating every $2\pi$, with values between $-1$ and $1$. It is an odd function, symmetric about the origin.
• Cosine is the same wave shifted left by $\frac{\pi}{2}$: it starts at its maximum 1 when $x = 0$, is 0 at $\frac{\pi}{2}$ and $-1$ at $\pi$. Period $2\pi$, range $[-1, 1]$, symmetric about the $y$-axis (even).
• Tangent is 0 at $x = 0$ and climbs without bound as $x$ approaches $\frac{\pi}{2}$, where it is undefined (a vertical asymptote). It repeats every $\pi$, not $2\pi$, and takes every real value.

Inverse functions

Because sine is not one-to-one, $\arcsin$ is defined by restricting sine to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So $\arcsin y$ is the one angle in that interval whose sine is $y$, for $y$ in $[-1, 1]$. Likewise $\arccos y$ lies in $[0, \pi]$, and $\arctan y$ lies strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ for every real $y$. Thus $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, not $\frac{7\pi}{6}$, and $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

Addition and double-angle formulas

$$\sin(a \pm b) = \sin a\cos b \pm \cos a\sin b \qquad \cos(a \pm b) = \cos a\cos b \mp \sin a\sin b$$

Setting $a = b = x$ gives the double-angle formulas:

$$\sin 2x = 2\sin x\cos x \qquad \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x = 2\cos^2 x - 1$$

Choose the version of $\cos 2x$ that leaves only one function in your equation. The addition formulas also give exact values such as $\cos 75° = \cos(45° + 30°)$.

Solving simple equations

An equation like $\sin x = \frac{1}{2}$ has infinitely many solutions: $x = \frac{\pi}{6} + 2k\pi$ and $x = \frac{5\pi}{6} + 2k\pi$ for every integer $k$. Two angles per turn share each sine (they are symmetric about the $y$-axis), and two share each cosine (symmetric about the $x$-axis). Questions usually restrict $x$ to an interval such as $[0, 2\pi)$; list the solutions in it and count carefully at the ends. Equations in $\sin^2 x$ and $\sin x$, or after a double-angle substitution, become quadratics in one function: solve for the function first, discard values outside $[-1, 1]$, then find the angles.

Simplifying also uses the identities: $\frac{1 - \cos^2 x}{\sin x\cos x} = \frac{\sin^2 x}{\sin x\cos x} = \tan x$, wherever the denominator is not zero.