Right triangle trigonometry
In a right triangle, the angles and the sides are linked. If you know one acute angle, you know the shape of the triangle. This page names the six ratios of sides that make this link. They are the six trigonometric functions.
The sides of a right triangle
A right triangle has one right angle (${90^\circ}$). The other two angles are acute, and they add up to ${90^\circ}$. The side across from the right angle is the hypotenuse. It is always the longest side. The other two sides are the legs.
Pick one of the acute angles, and call it ${\theta}$. We name the legs by where they sit compared to ${\theta}$:
- The opposite side is the leg across from ${\theta}$.
- The adjacent side is the leg next to ${\theta}$. ("Adjacent" means "next to.")
If you pick the other acute angle instead, the opposite and adjacent sides swap. The hypotenuse stays the same.
The six trigonometric functions
For an acute angle ${\theta}$ in a right triangle, the six trigonometric functions are these ratios:
$\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$$\csc\theta=\dfrac{\text{hyp}}{\text{opp}}$
$\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$$\sec\theta=\dfrac{\text{hyp}}{\text{adj}}$
$\tan\theta=\dfrac{\text{opp}}{\text{adj}}$$\cot\theta=\dfrac{\text{adj}}{\text{opp}}$
Here opp, adj, and hyp stand for the lengths of the opposite side, the adjacent side, and the hypotenuse. The names are short for sine, cosine, tangent, cosecant, secant, and cotangent.
Many students remember the first three with the made-up word SOH-CAH-TOA:
- SOH: Sine is Opposite over Hypotenuse.
- CAH: Cosine is Adjacent over Hypotenuse.
- TOA: Tangent is Opposite over Adjacent.
The last three are the first three flipped upside down. Two numbers whose product is ${1}$, like ${\dfrac{3}{5}}$ and ${\dfrac{5}{3}}$, are called reciprocals. So
${\csc\theta=\dfrac{1}{\sin\theta}}$
${\sec\theta=\dfrac{1}{\cos\theta}}$
${\cot\theta=\dfrac{1}{\tan\theta}}$
Why the ratios depend only on the angle
Take a right triangle and make it twice as big. Every side doubles, so every ratio stays the same. For example, a triangle with sides ${3}$, ${4}$, ${5}$ and one with sides ${6}$, ${8}$, ${10}$ have the same angles. In both, the side opposite ${\theta}$ over the hypotenuse is ${\dfrac{3}{5}=\dfrac{6}{10}}$.
Triangles with the same angles are called similar. Their sides are in the same ratios. So ${\sin\theta}$ depends only on the angle ${\theta}$, not on the size of the triangle. That is why we can call it a function of ${\theta}$.
Example 1: Find the six trigonometric functions of ${\theta}$ in the triangle below.
Solution:
From ${\theta}$, the opposite side is ${3}$, the adjacent side is ${4}$, and the hypotenuse is ${5}$.
$\sin\theta=\dfrac{3}{5}\qquad\csc\theta=\dfrac{5}{3}$
$\cos\theta=\dfrac{4}{5}\qquad\sec\theta=\dfrac{5}{4}$
$\tan\theta=\dfrac{3}{4}\qquad\cot\theta=\dfrac{4}{3}$
Example 2: The angle ${\theta}$ is acute, and ${\cos\theta=\dfrac{2}{3}}$. Find the other five trigonometric functions of ${\theta}$.
Solution:
Since ${\cos\theta=\dfrac{\text{adj}}{\text{hyp}}}$, draw a right triangle with adjacent side ${2}$ and hypotenuse ${3}$. (Any triangle with this ratio works, because the ratios depend only on the angle.)
Find the opposite side with the Pythagorean theorem, ${\text{opp}^2+\text{adj}^2=\text{hyp}^2}$:
$\begin{align*}\text{opp}^2+2^2&=3^2\\\text{opp}^2&=9-4=5\\\text{opp}&=\sqrt{5}\end{align*}$
Now read off the ratios:
$\sin\theta=\dfrac{\sqrt{5}}{3}\qquad\tan\theta=\dfrac{\sqrt{5}}{2}$
${\sec\theta=\dfrac{3}{2}}$
$\csc\theta=\dfrac{3}{\sqrt{5}}=\dfrac{3\sqrt{5}}{5}$
$\cot\theta=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}$
In the last two, we multiplied the top and bottom by ${\sqrt{5}}$. This removes the root from the bottom. It is called rationalizing the denominator.
Special triangles
Two right triangles come up so often that you should know their sides by heart.
The ${45^\circ}$-${45^\circ}$-${90^\circ}$ triangle. Its two acute angles are equal, so its two legs are equal. Let each leg be ${1}$. By the Pythagorean theorem, the hypotenuse is ${\sqrt{1^2+1^2}=\sqrt{2}}$.
The ${30^\circ}$-${60^\circ}$-${90^\circ}$ triangle. Start with an equilateral triangle (all three sides equal, all three angles ${60^\circ}$) with sides of length ${2}$. Cut it in half with a line from the top straight down. Each half is a right triangle with hypotenuse ${2}$, angles ${30^\circ}$ and ${60^\circ}$, and a short leg of ${1}$ (half of ${2}$). The long leg is ${\sqrt{2^2-1^2}=\sqrt{3}}$.
In this triangle, the short leg ${1}$ is across from the ${30^\circ}$ angle. The long leg ${\sqrt{3}}$ is across from the ${60^\circ}$ angle.
Reading the ratios from these two triangles gives the values below. Learn this table. You will use it in every chapter.
| ${\theta}$ | ${\sin\theta}$ | ${\cos\theta}$ | ${\tan\theta}$ |
|---|---|---|---|
| ${30^\circ=\dfrac{\pi}{6}}$ | ${\dfrac{1}{2}}$ | ${\dfrac{\sqrt{3}}{2}}$ | ${\dfrac{\sqrt{3}}{3}}$ |
| ${45^\circ=\dfrac{\pi}{4}}$ | ${\dfrac{\sqrt{2}}{2}}$ | ${\dfrac{\sqrt{2}}{2}}$ | ${1}$ |
| ${60^\circ=\dfrac{\pi}{3}}$ | ${\dfrac{\sqrt{3}}{2}}$ | ${\dfrac{1}{2}}$ | ${\sqrt{3}}$ |
For example, ${\tan 30^\circ=\dfrac{1}{\sqrt{3}}}$. Rationalizing gives ${\dfrac{\sqrt{3}}{3}}$.
A pattern helps you remember the sine column: ${\sin 30^\circ}$, ${\sin 45^\circ}$, ${\sin 60^\circ}$ are ${\dfrac{\sqrt{1}}{2}}$, ${\dfrac{\sqrt{2}}{2}}$, ${\dfrac{\sqrt{3}}{2}}$. The cosine column is the same list in reverse order.
Example 3: Find the exact value. "Exact" means no decimals: keep roots and fractions.
(a) $\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{6}-\tan\dfrac{\pi}{4}$
(b) ${\sin^2 45^\circ+\cos^2 30^\circ}$
Solution:
(a) From the table, ${\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}}$, ${\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}}$, and ${\tan\dfrac{\pi}{4}=1}$:
$\begin{align*}&\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{2}-1\\&=\dfrac{3}{4}-1=-\dfrac{1}{4}\end{align*}$
(b) The notation ${\sin^2\theta}$ means ${(\sin\theta)^2}$: find the sine first, then square it.
$\begin{align*}&\left(\dfrac{\sqrt{2}}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2\\&=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\end{align*}$
Cofunctions
Look again at the triangle at the top of this page. The two acute angles add up to ${90^\circ}$, so the top angle is ${90^\circ-\theta}$. The side opposite ${\theta}$ is adjacent to ${90^\circ-\theta}$, and the other way around. So the sine of one angle is the cosine of the other:
${\sin\theta=\cos(90^\circ-\theta)}$${\cos\theta=\sin(90^\circ-\theta)}$
${\tan\theta=\cot(90^\circ-\theta)}$${\cot\theta=\tan(90^\circ-\theta)}$
${\sec\theta=\csc(90^\circ-\theta)}$${\csc\theta=\sec(90^\circ-\theta)}$
Pairs like sine and cosine are called cofunctions. The "co" in cosine comes from "complementary angle." In radians, replace ${90^\circ}$ by ${\dfrac{\pi}{2}}$.
Example 4: Write each one using its cofunction.
(a) ${\sin 72^\circ}$
(b) ${\tan\dfrac{\pi}{12}}$
Solution:
(a) $\begin{align*}\sin 72^\circ&=\cos(90^\circ-72^\circ)\\&=\cos 18^\circ\end{align*}$
(b) Subtract the angle from ${\dfrac{\pi}{2}}$:
$\begin{align*}\dfrac{\pi}{2}-\dfrac{\pi}{12}&=\dfrac{6\pi}{12}-\dfrac{\pi}{12}\\&=\dfrac{5\pi}{12}\end{align*}$
So
${\tan\dfrac{\pi}{12}=\cot\dfrac{5\pi}{12}}$
Using a calculator
For most angles, there is no simple exact value. Use a calculator instead. Calculators have keys for sin, cos, and tan, but usually not for csc, sec, and cot. For those, use the reciprocals: for example, ${\sec\theta=\dfrac{1}{\cos\theta}}$.
Watch out: A calculator has two angle modes. Use degree mode when the angle is in degrees, and radian mode when it is in radians. In the wrong mode, you get a wrong answer with no warning. For example, ${\sin 30}$ in radian mode gives ${-0.988}$, not ${0.5}$.
Example 5: Use a calculator. Round to four decimal places.
(a) ${\sin 37^\circ}$ (b) ${\cot 50^\circ}$ (c) ${\sec 1.2}$
Solution:
(a) In degree mode: ${\sin 37^\circ\approx 0.6018}$
(b) In degree mode, find ${\tan 50^\circ\approx 1.1918}$. Then take the reciprocal:
$\cot 50^\circ=\dfrac{1}{\tan 50^\circ}\approx 0.8391$
(c) The angle ${1.2}$ has no degree sign, so it is in radians. In radian mode, find ${\cos 1.2\approx 0.3624}$. Then
$\sec 1.2=\dfrac{1}{\cos 1.2}\approx 2.7597$
Finding a missing side
If you know one side and one acute angle of a right triangle, you can find the other sides. Choose the trigonometric function that links the side you know with the side you want.
Example 6: Find ${x}$ and ${y}$ to two decimal places.
Solution:
From the ${35^\circ}$ angle, ${y}$ is the opposite side, ${x}$ is the adjacent side, and ${12}$ is the hypotenuse.
To find ${y}$, use sine (opposite and hypotenuse):
$\begin{align*}\sin 35^\circ&=\dfrac{y}{12}\\y&=12\sin 35^\circ\approx 6.88\end{align*}$
To find ${x}$, use cosine (adjacent and hypotenuse):
$\begin{align*}\cos 35^\circ&=\dfrac{x}{12}\\x&=12\cos 35^\circ\approx 9.83\end{align*}$
Check: By the Pythagorean theorem, ${x^2+y^2}$ should be ${12^2=144}$. Indeed, ${9.83^2+6.88^2\approx 144}$.
The lesson Solving right triangles uses this idea to find heights and distances.
Summary
- For an acute angle ${\theta}$ of a right triangle: ${\sin\theta=\dfrac{\text{opp}}{\text{hyp}}}$, ${\cos\theta=\dfrac{\text{adj}}{\text{hyp}}}$, ${\tan\theta=\dfrac{\text{opp}}{\text{adj}}}$ (SOH-CAH-TOA).
- Cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent.
- The ratios depend only on the angle, not on the size of the triangle.
- Learn the special triangles: ${1}$, ${1}$, ${\sqrt{2}}$ and ${1}$, ${\sqrt{3}}$, ${2}$, and the table of values for ${30^\circ}$, ${45^\circ}$, and ${60^\circ}$.
- Cofunctions: ${\sin\theta=\cos(90^\circ-\theta)}$, and so on.
- On a calculator, check the mode: degrees or radians.
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