Sine, Cosine, and Tangent

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Learning goals

Identify opposite and adjacent legs relative to a selected acute angle
Prove that AA similarity makes side ratios independent of triangle size
Evaluate sine, cosine, and tangent from exact right-triangle lengths
Connect complementary-angle ratios and derive exact values for special angles
Justify the quotient and Pythagorean identities for acute angles

Choose an angle before naming a leg

In this lesson, every reference angle is acute, between $0^\circ$ and $90^\circ$ , in a nondegenerate right triangle. We use degrees throughout.

A reference angle is the acute angle selected for comparing sides. The opposite leg lies across from it. The adjacent leg meets it and the right angle. The hypotenuse also meets the reference angle, but it is excluded from the term adjacent leg.

For example, take $\triangle ABC$ , right at $C$ , with lengths

\begin{aligned} BC&=10\\ CA&=24\\ AB&=26. \end{aligned}

Choose angle $A$ , labeled $\theta$ (the Greek letter theta).

The marked acute angle determines which leg is opposite and which is adjacent.

For $\theta$ at $A$ , the opposite leg is $BC$ , the adjacent leg is $CA$ , and the hypotenuse is $AB$ . Choosing $B$ instead exchanges the two leg roles. The hypotenuse stays opposite the right angle, so it does not change.

Check your understanding

Triangle ABC is right at C. If B is selected as the acute reference angle, which side is its adjacent leg?

A AB B CA C BC D Both AB and BC

Why the ratios depend on the angle

Two right triangles with the same selected acute angle have the same three angle measures. The AA criterion therefore makes them similar. Their sizes can differ while their matching side ratios stay fixed.

An acute angle fixes the three side ratios#

First compare a triangle with opposite, adjacent, and hypotenuse lengths $6$ , $8$ , and $10$ with one whose corresponding lengths are $15$ , $20$ , and $25$ . Both are right triangles, and the second is a dilation of the first with factor $2.5$ . Their chosen acute angles match.

The panels use different drawing scales. Compare the labeled side lengths, with the selected angle at A.

The second panel uses its own drawing scale; the corresponding selected angle is at D.

Each ratio cancels that common factor:

\begin{aligned} \frac{15}{25}&=\frac6{10}\\ \frac{20}{25}&=\frac8{10}\\ \frac{15}{20}&=\frac68. \end{aligned}

Now consider any two nondegenerate right triangles with the same acute reference angle. They have a matching right angle and the given matching acute angle, so AA proves similarity. Write $a,b,c$ for the first triangle’s opposite leg, adjacent leg, and hypotenuse. If $k>0$ is the scale factor to the second triangle, its corresponding lengths are $ka,kb,kc$ .

\begin{aligned} \frac{ka}{kc}&=\frac ac\\ \frac{kb}{kc}&=\frac bc\\ \frac{ka}{kb}&=\frac ab. \end{aligned}

The positive scale factor cancels in each ratio. Nothing depended on $2.5$ or on the particular side lengths: all three ratios are fixed by the acute angle.

Keep the angle, change the size

Reference A: 35°

Scroll sideways to explore the triangle.

Opposite leg BC ≈ 3.441 units
Adjacent leg CA ≈ 4.915 units
Hypotenuse AB = 6.000 units

Angle A Size AB Reference Turn

The selected acute angle A measures 35°. The opposite leg BC is about 3.441 units long. The adjacent leg CA is about 4.915 units long. The hypotenuse AB is 6.000 units long. Sine of A, the ratio BC/AB, is about 0.574. Cosine of A, the ratio CA/AB, is about 0.819. Tangent of A, the ratio BC/CA, is about 0.700. The drawing is turned 0° counterclockwise from its starting pose.

Change size with angle A fixed, then compare the ratios as the angle or reference changes.

Check your understanding

Two right triangles share the same acute reference angle. A dilation doubles every side. What happens to opposite leg divided by hypotenuse?

A It doubles B It is halved C It is squared D It stays unchanged

Give the three ratios names

For an acute angle $\theta$ in a right triangle, define its sine, cosine, and tangent by

\begin{aligned} \sin\theta&=\frac{\text{opposite leg}}{\text{hypotenuse}}\\ \cos\theta&=\frac{\text{adjacent leg}}{\text{hypotenuse}}\\ \tan\theta&=\frac{\text{opposite leg}}{\text{adjacent leg}}. \end{aligned}

Here $\sin$ , $\cos$ , and $\tan$ abbreviate the three names. They refer to ratios, not lengths. The units cancel when both lengths use the same unit.

The mnemonic SOH CAH TOA records these definitions: sine uses opposite over hypotenuse, cosine uses adjacent over hypotenuse, and tangent uses opposite over adjacent. It works only after the reference angle and side roles have been identified.

Worked example 1 Read three ratios in a turned triangle

Problem. Triangle $LMN$ is right at $M$ . Its side lengths are

\begin{aligned} LM&=24\text{ cm}\\ MN&=7\text{ cm}\\ LN&=25\text{ cm}. \end{aligned}

Use $L$ as the reference angle. Find its sine, cosine, and tangent exactly.

The marked angle at L is the selected reference angle.

Solution. Opposite $L$ is $MN$ , adjacent to $L$ is leg $LM$ , and the hypotenuse is $LN$ .

\begin{aligned} \sin L&=\frac7{25}\\ \cos L&=\frac{24}{25}\\ \tan L&=\frac7{24}. \end{aligned}

Turning the triangle changes none of these side roles relative to $L$ .

Because a hypotenuse is longer than either positive leg, sine and cosine of an acute angle are strictly between $0$ and $1$ . Tangent is positive, but it can exceed $1$ when the opposite leg is longer than the adjacent leg.

Switch to the other acute angle

The two acute angles of a right triangle are complementary: their measures add to $90^\circ$ . Switching between them exchanges opposite and adjacent legs.

Worked example 2 Use the other reference angle

Problem. Triangle $ABC$ is right at $C$ . The acute angles at $A$ and $B$ are labeled $\theta$ and $\phi$ . Its lengths are

\begin{aligned} BC&=4\\ CA&=\sqrt{65}\\ AB&=9. \end{aligned}

(a) Identify the opposite leg, adjacent leg, and hypotenuse for $\phi$ .

(b) Find all three ratios for $\phi$ , and compare $\cos\phi$ with $\sin\theta$ .

The two angle markings distinguish the selected acute angles.

Solution. (a) For $\phi$ at $B$ , the opposite leg is $CA$ , the adjacent leg is $BC$ , and the hypotenuse is still $AB$ .

(b) Substitute those roles into the definitions.

\begin{aligned} \sin\phi&=\frac{\sqrt{65}}9\\ \cos\phi&=\frac49\\ \tan\phi&=\frac{\sqrt{65}}4. \end{aligned}

For $\theta$ , the opposite leg is $BC$ , so $\sin\theta=4/9$ . The same side ratio gives $\cos\phi$ .

In any right triangle, a leg opposite one acute angle is adjacent to the other. Thus, for acute $\theta$ and its acute complement,

\begin{aligned} \sin(90^\circ-\theta)&=\cos\theta\\ \cos(90^\circ-\theta)&=\sin\theta. \end{aligned}

Tangent swaps its two positive legs, so

\tan(90^\circ-\theta)=\frac1{\tan\theta}.

For example, the tangent of $\theta$ in the worked triangle is $4/\sqrt{65}$ , while the tangent of $\phi$ is $\sqrt{65}/4$ . The two values multiply to $1$ . Sine and cosine exchange; tangent becomes its reciprocal.

Check your understanding

Angles θ and φ are acute and complementary. Given that $\sin\theta=\frac27$ , what is $\cos\phi$ ?

\frac27

\frac72

\frac57

\frac{\sqrt{45}}7

Exact values from the special triangles

The previous lesson derived the two special side ratios. Their trig values come from dividing those already justified lengths.

In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, use short leg $1$ , long leg $\sqrt3$ , and hypotenuse $2$ .

The side lengths belong to the established half-equilateral ratio.

Relative to $30^\circ$ , the definitions give

\begin{aligned} \sin30^\circ&=\frac12\\ \cos30^\circ&=\frac{\sqrt3}{2}\\ \tan30^\circ&=\frac1{\sqrt3}\\ \tan30^\circ&=\frac{\sqrt3}{3}. \end{aligned}

Switching to $60^\circ$ exchanges the legs, so its sine and cosine exchange those values and its tangent is $\sqrt3$ .

For $45^\circ$ , the established side lengths can be $1$ , $1$ , and $\sqrt2$ .

The equal-leg shape supplies the 45-degree ratios.

Both sine and cosine are $1/\sqrt2$ , which rationalizes to $\sqrt2/2$ . Tangent is $1/1$ , or $1$ . The complete exact values are

Acute angle	Sine	Cosine	Tangent
$30^\circ$	$\frac12$	$\frac{\sqrt3}{2}$	$\frac{\sqrt3}{3}$
$45^\circ$	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	$1$
$60^\circ$	$\frac{\sqrt3}{2}$	$\frac12$	$\sqrt3$

Worked example 3 Combine two exact values

Problem. Evaluate the following expression exactly, with no decimal approximation:

\frac{\sin45^\circ}{\tan30^\circ}.

Solution. Substitute the values derived from the two special triangles.

\begin{aligned} \frac{\sin45^\circ}{\tan30^\circ}&=\frac{\sqrt2/2}{\sqrt3/3}\\ &=\frac{3\sqrt2}{2\sqrt3}\\ &=\frac{\sqrt6}{2}. \end{aligned}

Dividing the two ratios is ordinary fraction arithmetic.

Two identities from the same sides

An identity is an equality that holds for every allowed input. Here the allowed angles are acute angles in nondegenerate right triangles.

The notation $\sin^2\theta$ means $(\sin\theta)^2$ , the square of the sine value. It does not mean $\sin(\theta^2)$ . Cosine squared and tangent squared use the same convention.

The quotient and Pythagorean identities for acute angles#

Take a right triangle with opposite leg $5$ , adjacent leg $12$ , and hypotenuse $13$ , relative to $\theta$ .

The same three side lengths are used in both identities.

Dividing sine by cosine cancels their common hypotenuse.

\begin{aligned} \frac{5/13}{12/13}&=\frac5{12}\\ \frac{\sin\theta}{\cos\theta}&=\tan\theta. \end{aligned}

Squaring sine and cosine and adding gives

\begin{aligned} \left(\frac5{13}\right)^2+\left(\frac{12}{13}\right)^2&=\frac{25+144}{169}\\ \left(\frac5{13}\right)^2+\left(\frac{12}{13}\right)^2&=1. \end{aligned}

For a general right triangle $ABC$ , let $\theta$ be at $A$ , with $a=BC$ , $b=CA$ , and $c=AB$ . These lengths are positive. The same fraction cancellation gives

\begin{aligned} \frac{\sin\theta}{\cos\theta}&=\frac{a/c}{b/c}\\ \frac{\sin\theta}{\cos\theta}&=\frac ab\\ \tan\theta&=\frac{\sin\theta}{\cos\theta}. \end{aligned}

Division is valid because the adjacent leg $b$ is positive, so $\cos\theta$ is positive. The Pythagorean theorem gives

a^2+b^2=c^2.

Therefore

\begin{aligned} \sin^2\theta+\cos^2\theta&=\frac{a^2+b^2}{c^2}\\ \sin^2\theta+\cos^2\theta&=1. \end{aligned}

The numerical triangle only illustrated the two cancellations. AA makes these definitions valid for any triangle with the chosen angle, and the two arguments apply to every acute $\theta$ .

Worked example 4 Recover related ratios

Problem. An acute angle $\theta$ satisfies

\cos\theta=\frac38.

Find $\sin\theta$ and $\tan\theta$ exactly.

Solution. The squared-value identity gives

\begin{aligned} \sin^2\theta&=1-\left(\frac38\right)^2\\ \sin^2\theta&=\frac{55}{64}. \end{aligned}

The sine is positive for an acute angle, so

\sin\theta=\frac{\sqrt{55}}8.

Now divide sine by cosine.

\begin{aligned} \tan\theta&=\frac{\sqrt{55}/8}{3/8}\\ \tan\theta&=\frac{\sqrt{55}}3. \end{aligned}

A possible triangle with these ratios, drawn after the calculation.

Check your understanding

For an acute θ, which statement is correct?

\sin\theta+\cos\theta=1

\sin^2\theta+\cos^2\theta=1

\tan\theta=\sin\theta\cos\theta

\cos\theta

can be zero

Sine, Cosine, and Tangent

Learning goals

Choose an angle before naming a leg

Why the ratios depend on the angle

Give the three ratios names

Switch to the other acute angle

Exact values from the special triangles

Two identities from the same sides

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress