Trigonometric functions of any angle
In a right triangle, the angles are less than ${90^\circ}$. But angles can be any size: ${150^\circ}$, ${300^\circ}$, or even negative. This page defines the six trigonometric functions for every angle. It uses a point on the terminal side of the angle instead of a triangle.
The definitions
Put an angle ${\theta}$ in standard position. Choose any point ${P(x,y)}$ on its terminal side, other than the origin. Let ${r}$ be the distance from the origin to ${P}$. By the distance formula,
${r=\sqrt{x^2+y^2}}$
The distance ${r}$ is always positive. But ${x}$ and ${y}$ can be positive or negative, depending on the quadrant.
The six trigonometric functions of ${\theta}$ are
${\sin\theta=\dfrac{y}{r}}$${\csc\theta=\dfrac{r}{y}}$
${\cos\theta=\dfrac{x}{r}}$${\sec\theta=\dfrac{r}{x}}$
${\tan\theta=\dfrac{y}{x}}$${\cot\theta=\dfrac{x}{y}}$
A function is undefined when its denominator is ${0}$. So ${\tan\theta}$ and ${\sec\theta}$ are undefined when ${x=0}$, and ${\cot\theta}$ and ${\csc\theta}$ are undefined when ${y=0}$.
Why these match the old definitions. If ${\theta}$ is acute, ${P}$ is in quadrant I. The dashed line makes a right triangle. Its side adjacent to ${\theta}$ is ${x}$, its opposite side is ${y}$, and its hypotenuse is ${r}$. So ${\dfrac{y}{r}}$ is opposite over hypotenuse, just as before.
As with triangles, the answer does not depend on which point ${P}$ you choose. A point twice as far out has ${x}$, ${y}$, and ${r}$ all doubled, so the ratios stay the same.
Example 1: The terminal side of ${\theta}$ passes through the point ${(-3,4)}$. Find the six trigonometric functions of ${\theta}$.
Solution:
Here ${x=-3}$ and ${y=4}$. Find ${r}$:
${r=\sqrt{(-3)^2+4^2}=\sqrt{25}=5}$
Now use the definitions:
${\sin\theta=\dfrac{4}{5}}$${\csc\theta=\dfrac{5}{4}}$
${\cos\theta=-\dfrac{3}{5}}$${\sec\theta=-\dfrac{5}{3}}$
${\tan\theta=-\dfrac{4}{3}}$${\cot\theta=-\dfrac{3}{4}}$
The functions that use ${x}$ are negative here, because ${x}$ is negative in quadrant II.
Quadrantal angles
For a quadrantal angle, the terminal side lies on an axis. Choose the point ${P}$ at distance ${r=1}$ from the origin. Then ${P}$ is ${(1,0)}$, ${(0,1)}$, ${(-1,0)}$, or ${(0,-1)}$.
Example 2: Find ${\sin\theta}$, ${\cos\theta}$, and ${\tan\theta}$ for ${\theta=0}$, ${90^\circ}$, ${180^\circ}$, and ${270^\circ}$.
Solution:
For ${\theta=90^\circ}$, the terminal side is the positive ${y}$-axis. Take ${P(0,1)}$, so ${x=0}$, ${y=1}$, and ${r=1}$:
${\sin 90^\circ=\dfrac{1}{1}=1}$${\cos 90^\circ=\dfrac{0}{1}=0}$
${\tan 90^\circ=\dfrac{1}{0}}$, which is undefined.
The other angles work the same way, with ${P(1,0)}$, ${P(-1,0)}$, and ${P(0,-1)}$:
| ${\theta}$ | ${\sin\theta}$ | ${\cos\theta}$ | ${\tan\theta}$ |
|---|---|---|---|
| ${0}$ | ${0}$ | ${1}$ | ${0}$ |
| ${90^\circ=\dfrac{\pi}{2}}$ | ${1}$ | ${0}$ | undefined |
| ${180^\circ=\pi}$ | ${0}$ | ${-1}$ | ${0}$ |
| ${270^\circ=\dfrac{3\pi}{2}}$ | ${-1}$ | ${0}$ | undefined |
Signs in each quadrant
Since ${r}$ is always positive, the sign of each function comes from the signs of ${x}$ and ${y}$. For example, ${\sin\theta=\dfrac{y}{r}}$ is positive when ${y}$ is positive, that is, in quadrants I and II.
Each function has the same sign as its reciprocal. Working through all four quadrants gives this picture:
A sentence helps you remember it. Going counterclockwise from quadrant I: "All Students Take Calculus." The first letters stand for All, Sine, Tangent, Cosine.
Example 3: In which quadrant does ${\theta}$ lie if ${\sin\theta<0}$ and ${\tan\theta>0}$?
Solution:
Sine is negative in quadrants III and IV. Tangent is positive in quadrants I and III. Both are true only in quadrant III.
Example 4: Suppose ${\tan\theta=-\dfrac{8}{15}}$ and ${\cos\theta>0}$. Find the other five trigonometric functions of ${\theta}$.
Solution:
Step 1: Find the quadrant. Tangent is negative in quadrants II and IV. Cosine is positive in quadrants I and IV. So ${\theta}$ is in quadrant IV, where ${x>0}$ and ${y<0}$.
Step 2: Choose a point. Since ${\tan\theta=\dfrac{y}{x}=\dfrac{-8}{15}}$, take ${x=15}$ and ${y=-8}$. Then
$\begin{align*}r&=\sqrt{15^2+(-8)^2}\\&=\sqrt{289}=17\end{align*}$
Step 3: Use the definitions.
${\sin\theta=-\dfrac{8}{17}}$${\csc\theta=-\dfrac{17}{8}}$
${\cos\theta=\dfrac{15}{17}}$${\sec\theta=\dfrac{17}{15}}$
${\cot\theta=-\dfrac{15}{8}}$
Reference angles
The reference angle of ${\theta}$ is the acute angle between the terminal side of ${\theta}$ and the ${x}$-axis. We write it ${\theta'}$ (read "theta prime"). It is always between ${0^\circ}$ and ${90^\circ}$.
For ${\theta}$ between ${0^\circ}$ and ${360^\circ}$, the figures give these rules:
| ${\theta}$ in | ${\theta'}$ (degrees) | ${\theta'}$ (radians) |
|---|---|---|
| Quadrant II | ${180^\circ-\theta}$ | ${\pi-\theta}$ |
| Quadrant III | ${\theta-180^\circ}$ | ${\theta-\pi}$ |
| Quadrant IV | ${360^\circ-\theta}$ | ${2\pi-\theta}$ |
If ${\theta}$ is in quadrant I, then ${\theta'=\theta}$. If ${\theta}$ is larger than ${360^\circ}$ or negative, first find a coterminal angle between ${0^\circ}$ and ${360^\circ}$.
Why reference angles help
Reflect the point ${P(x,y)}$ across one or both axes into quadrant I. Its coordinates keep their sizes but may change signs. So the function values of ${\theta}$ and ${\theta'}$ are the same, except perhaps for the sign:
${\sin\theta=\pm\sin\theta'}$, and the same for the other five functions.
The sign (${+}$ or ${-}$) is the sign of the function in the quadrant of ${\theta}$.
This gives a three-step method to find the value of a trigonometric function of any angle:
- Find the reference angle ${\theta'}$.
- Find the value of the function at ${\theta'}$.
- Choose the sign from the quadrant of ${\theta}$.
Example 5: Find the exact value.
(a) ${\sin 150^\circ}$ (b) ${\cos 225^\circ}$
(c) ${\tan\dfrac{5\pi}{3}}$ (d) ${\sec(-120^\circ)}$
Solution:
(a) ${150^\circ}$ is in quadrant II, so ${\theta'=180^\circ-150^\circ=30^\circ}$. Sine is positive in quadrant II:
$\sin 150^\circ=+\sin 30^\circ=\dfrac{1}{2}$
(b) ${225^\circ}$ is in quadrant III, so ${\theta'=225^\circ-180^\circ=45^\circ}$. Cosine is negative in quadrant III:
$\cos 225^\circ=-\cos 45^\circ=-\dfrac{\sqrt{2}}{2}$
(c) ${\dfrac{5\pi}{3}}$ is between ${\dfrac{3\pi}{2}}$ and ${2\pi}$, so it is in quadrant IV. Its reference angle is
$\theta'=2\pi-\dfrac{5\pi}{3}=\dfrac{\pi}{3}$
Tangent is negative in quadrant IV:
$\tan\dfrac{5\pi}{3}=-\tan\dfrac{\pi}{3}=-\sqrt{3}$
(d) Add ${360^\circ}$ to get a coterminal angle: ${-120^\circ+360^\circ=240^\circ}$. This is in quadrant III, so ${\theta'=240^\circ-180^\circ=60^\circ}$. Cosine is negative in quadrant III, so
$\cos(-120^\circ)=-\cos 60^\circ=-\dfrac{1}{2}$
Secant is the reciprocal of cosine:
$\sec(-120^\circ)=\dfrac{1}{-\dfrac{1}{2}}=-2$
A calculator also gives values for any angle. For example, in degree mode, ${\sin 200^\circ\approx -0.3420}$. But a calculator gives decimals, not exact values like ${-\dfrac{\sqrt{2}}{2}}$.
The fundamental identities
An identity is an equation that is true for every value of the variable where both sides are defined. The definitions give several identities at once.
Reciprocal identities. These come straight from the definitions:
${\csc\theta=\dfrac{1}{\sin\theta}}$${\sec\theta=\dfrac{1}{\cos\theta}}$${\cot\theta=\dfrac{1}{\tan\theta}}$
Quotient identities. Divide ${\sin\theta=\dfrac{y}{r}}$ by ${\cos\theta=\dfrac{x}{r}}$. The ${r}$'s cancel, and you get ${\dfrac{y}{x}}$, which is ${\tan\theta}$:
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$
Pythagorean identities. Since ${r=\sqrt{x^2+y^2}}$, we have ${x^2+y^2=r^2}$. Divide both sides by ${r^2}$:
$\left(\dfrac{x}{r}\right)^2+\left(\dfrac{y}{r}\right)^2=1$
This says ${\cos^2\theta+\sin^2\theta=1}$. Dividing ${x^2+y^2=r^2}$ by ${x^2}$ or by ${y^2}$ instead gives two more:
${\sin^2\theta+\cos^2\theta=1}$
${1+\tan^2\theta=\sec^2\theta}$
${1+\cot^2\theta=\csc^2\theta}$
Example 6: Suppose ${\sin\theta=\dfrac{2}{3}}$ and ${\theta}$ is in quadrant II. Find ${\cos\theta}$ and ${\tan\theta}$.
Solution:
Use ${\sin^2\theta+\cos^2\theta=1}$ to find ${\cos\theta}$. Here $\sin^2\theta=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}$, so
$\begin{align*}\dfrac{4}{9}+\cos^2\theta&=1\\\cos^2\theta&=1-\dfrac{4}{9}\\&=\dfrac{5}{9}\\\cos\theta&=\pm\dfrac{\sqrt{5}}{3}\end{align*}$
Cosine is negative in quadrant II, so ${\cos\theta=-\dfrac{\sqrt{5}}{3}}$.
Now use the quotient identity:
$\begin{align*}\tan\theta&=\dfrac{\sin\theta}{\cos\theta}=\dfrac{2/3}{-\sqrt{5}/3}\\&=-\dfrac{2}{\sqrt{5}}=-\dfrac{2\sqrt{5}}{5}\end{align*}$
Watch out: Taking a square root gives two answers, ${+}$ and ${-}$. Always use the quadrant to choose the right sign.
Summary
- For a point ${P(x,y)}$ on the terminal side of ${\theta}$, with ${r=\sqrt{x^2+y^2}}$: ${\sin\theta=\dfrac{y}{r}}$, ${\cos\theta=\dfrac{x}{r}}$, ${\tan\theta=\dfrac{y}{x}}$.
- Signs by quadrant: All, Sine, Tangent, Cosine are positive in quadrants I, II, III, IV.
- The reference angle ${\theta'}$ is the acute angle between the terminal side and the ${x}$-axis. A function of ${\theta}$ equals the same function of ${\theta'}$, with the sign from the quadrant.
- Identities: ${\tan\theta=\dfrac{\sin\theta}{\cos\theta}}$ and ${\sin^2\theta+\cos^2\theta=1}$, plus the reciprocal identities.