Math Core

If you know one acute angle of a right triangle, you already know its shape, even before you know its size. Trigonometry turns that shape into numbers: six ratios of side lengths that let you move back and forth between angles and sides.

Naming the sides from an angle

Every right triangle has one side that never changes its name: the hypotenuse, the longest side, across from the right angle. The other two sides are the legs, and their names depend on which acute angle you are standing at.

The opposite side is the leg across from your angle. It does not touch the angle.
The adjacent side is the leg that forms one side of your angle (along with the hypotenuse).

The labels opposite and adjacent are measured from the angle θ. The hypotenuse is always across from the right angle.

If you switch to the other acute angle, the two legs swap names: the side that was opposite becomes adjacent, and the other way around. So always ask, "opposite and adjacent to which angle?"

Why ratios depend only on the angle

Draw two right triangles that both have a $35^\circ$ angle. Their angles match ( $35^\circ$ , $55^\circ$ , $90^\circ$ ), so the triangles are similar by AA. In similar triangles, corresponding sides are proportional, which means the ratio $\dfrac{\text{opposite}}{\text{hypotenuse}}$ is the same in both, no matter how big they are. That ratio is a property of the angle $35^\circ$ alone. This is what makes it possible to give each ratio a name and treat it as a function of the angle.

The three primary ratios

Definition

Sine, cosine and tangent

For an acute angle $\theta$ in a right triangle:

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}

The memory aid SOH-CAH-TOA packs all three together: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

Because the hypotenuse is the longest side, $\sin\theta$ and $\cos\theta$ are always between $0$ and $1$ for an acute angle. Tangent has no such limit: it is less than $1$ when the opposite leg is shorter than the adjacent leg, and greater than $1$ when it is longer.

The three reciprocal ratios

Flip each primary ratio upside down and you get the other three trigonometric ratios.

Definition

Cosecant, secant and cotangent

\csc\theta = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{1}{\tan\theta}

Notice the pairing: cosecant goes with sine, and secant goes with cosine. Each "co" name pairs with a name that lacks it. Cotangent pairs with tangent. Since sine and cosine are at most $1$ , cosecant and secant are always at least $1$ for an acute angle.

Worked example: All six ratios from a diagram

Find all six trigonometric ratios of $\angle A$ .

Right triangle ABC with legs 15 and 8 and hypotenuse 17.

From $A$ : the opposite side is $BC = 8$ , the adjacent side is $AC = 15$ , and the hypotenuse is $AB = 17$ .

\begin{aligned} \sin A &= \tfrac{8}{17} & \cos A &= \tfrac{15}{17} & \tan A &= \tfrac{8}{15} \\ \csc A &= \tfrac{17}{8} & \sec A &= \tfrac{17}{15} & \cot A &= \tfrac{15}{8} \end{aligned}

Check: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ , so the side lengths really do form a right triangle.

Finding the ratios from just one of them

If you know one ratio, you can build a triangle that has it and fill in the missing side with the Pythagorean theorem.

Worked example: Given the tangent

$\theta$ is an acute angle with $\tan\theta = \dfrac{7}{24}$ . Find the other five ratios.

Draw a right triangle with opposite leg $7$ and adjacent leg $24$ . The hypotenuse is

\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25.

So $\sin\theta = \dfrac{7}{25}$ , $\cos\theta = \dfrac{24}{25}$ , $\csc\theta = \dfrac{25}{7}$ , $\sec\theta = \dfrac{25}{24}$ and $\cot\theta = \dfrac{24}{7}$ .

The actual triangle might be $14$ , $48$ , $50$ or any other multiple of $7$ , $24$ , $25$ . The ratios come out the same, which is exactly the similarity argument from before.

Worked example: When the hypotenuse is a radical

In right triangle $ABC$ with right angle $C$ , $AC = 3$ and $BC = 2$ . Find $\sin A$ , $\cos A$ and $\csc A$ exactly.

Right triangle ABC with legs 3 and 2.

The hypotenuse is $AB = \sqrt{3^2 + 2^2} = \sqrt{13}$ . From $A$ , the opposite side is $2$ and the adjacent side is $3$ :

\sin A = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}, \qquad \cos A = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}, \qquad \csc A = \frac{\sqrt{13}}{2}.

Multiplying the top and bottom by $\sqrt{13}$ rationalizes the denominator, the usual way to write an exact answer.

Common mistake

Opposite and adjacent are always relative to the angle you are working with. In the triangle above, $\sin A = \dfrac{2\sqrt{13}}{13}$ but $\sin B = \dfrac{3\sqrt{13}}{13}$ . Before you write any ratio, put your finger on the angle and name the three sides from there.

Cofunctions

Look again at the $8$ - $15$ - $17$ triangle. From $B$ , the opposite side is $15$ , so $\sin B = \dfrac{15}{17}$ , which is exactly $\cos A$ . That is no accident. The two acute angles of a right triangle add up to $90^\circ$ (they are complementary), and the side opposite one of them is adjacent to the other.

Cofunction identities

For any acute angle $\theta$ (in degrees):

\begin{aligned} \sin\theta &= \cos(90^\circ - \theta) & \tan\theta &= \cot(90^\circ - \theta) & \sec\theta &= \csc(90^\circ - \theta) \\ \cos\theta &= \sin(90^\circ - \theta) & \cot\theta &= \tan(90^\circ - \theta) & \csc\theta &= \sec(90^\circ - \theta) \end{aligned}

A trig ratio of an angle equals its cofunction of the complementary angle. The name "cosine" literally means "the sine of the complement."

Worked example: Using a cofunction to solve for x

Find $x$ if $\cos(2x)^\circ = \sin(x + 30)^\circ$ , where both angles are acute.

Sine and cosine are cofunctions, so the two angles must be complementary:

\begin{aligned} 2x + (x + 30) &= 90 \\ 3x &= 60 \\ x &= 20 \end{aligned}

Check: the angles are $2(20) = 40^\circ$ and $20 + 30 = 50^\circ$ , which add to $90^\circ$ . Indeed $\cos 40^\circ = \sin 50^\circ$ .

Tip

Quick check for any set of answers: $\sin\theta$ times $\csc\theta$ must be $1$ , and so must $\cos\theta \cdot \sec\theta$ and $\tan\theta \cdot \cot\theta$ . Also, $\tan\theta$ should equal $\dfrac{\sin\theta}{\cos\theta}$ , because the hypotenuses cancel.