Why do sine and cosine of an acute angle always come out less than 1?

Both ratios have the hypotenuse in the denominator, and the hypotenuse is the longest side of a right triangle. A fraction whose numerator is a leg and whose denominator is the hypotenuse must be less than 1. If you ever compute =13/5 for an acute angle, you have flipped a fraction or mislabeled the hypotenuse. Tangent is different: it compares two legs, so it can be less than 1, equal to 1 (at 45^), or arbitrarily large.

Does the size of the triangle change the value of a trigonometric ratio?

No. Any two right triangles sharing an acute angle are similar by AA, so their corresponding sides are proportional and every side-to-side ratio matches. Doubling the triangle doubles both the numerator and the denominator, leaving the fraction unchanged. This is exactly why a calculator can report a single value for 41^ without asking you how big your triangle is.

What is the difference between and ^-1 on my calculator?

The key takes an angle in and returns a ratio. The ^-1 key (inverse sine) takes a ratio in and returns the angle. In this lesson you mostly go in the first direction — angle to ratio — and set up ratios from given side lengths. Finding the angle from a ratio is the focus of the next lesson on solving right triangles. Note that ^-1 does not mean 1/.

Is there a cofunction rule for tangent like the one for sine and cosine?

Yes, but it is a reciprocal rather than an equality. Since swapping to the complementary angle trades the opposite and adjacent legs, (90^-)=adjacentopposite=1/. That reciprocal is called cotangent. So 30^ and 60^ are not equal — their product is 1.

GEOM-7.3

Trigonometric Ratios: Sine, Cosine & Tangent

Learn how sine, cosine, and tangent are defined in right triangles, why similarity makes them depend only on the angle, and how sine and cosine link complementary angles.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Trigonometric Ratios: Sine, Cosine & Tangent, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Every right triangle with a

37^\circ

angle — whether it fits on your desk or spans a canyon — has the same shape. Similar triangles have proportional sides, so the ratio of any two sides is locked in by the angle alone. That single fact is what makes trigonometry possible: instead of measuring a new triangle every time, you can look up (or compute) one number for the angle and use it forever.

In this lesson you will define the three basic trigonometric ratios — sine, cosine, and tangent — learn to label sides correctly as opposite, adjacent, and hypotenuse from the point of view of a chosen acute angle, and see the similarity argument that guarantees these ratios are well defined. You will also discover why

\sin 25^\circ

and

\cos 65^\circ

are exactly the same number, a relationship that comes straight from the fact that the two acute angles of a right triangle are complementary.

Labeling Sides: Opposite, Adjacent, and Hypotenuse

Before you can write a ratio, you have to name the sides — and two of the three names change depending on which acute angle you are standing at.

The hypotenuse is always the side across from the right angle; it never changes and it is always the longest side. Once you pick an acute angle (call it

\theta

), the opposite leg is the leg that does not touch

\theta

, and the adjacent leg is the leg that does touch

\theta

but is not the hypotenuse.

With those names, the three ratios are

\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 device SOH-CAH-TOA encodes exactly this. But a memory device is useless if you mislabel the sides, and that is where most early mistakes happen.

The biggest trap: students label the legs once and then keep those labels when the problem switches to the other acute angle. In a triangle with legs of 5 and 12, the leg of length 5 is opposite one acute angle and adjacent to the other. Whenever the angle changes, re-label from scratch.

A second trap is calling a leg the hypotenuse because it looks long or because the triangle is drawn rotated. Find the right angle first, then go straight across from it. If the triangle is tilted or flipped, redraw it in your notebook in a familiar orientation before writing any ratio.

One more habit worth building: write the ratio symbolically before plugging in numbers, as in

\tan A=\frac{BC}{AC}

. It forces you to commit to the labels.

Why the Ratios Depend Only on the Angle

Here is the question that makes trigonometry legitimate: if two different people draw right triangles that both contain a

40^\circ

angle but use completely different sizes, do they get the same value for

\sin 40^\circ

?

Yes — and the reason is the AA Similarity criterion. Suppose triangle

ABC

and triangle

DEF

each have a right angle and each have an acute angle measuring

40^\circ

. Two pairs of angles are congruent, so

\triangle ABC\sim\triangle DEF

. Similar triangles have proportional corresponding sides, which means

\frac{BC}{AB}=\frac{EF}{DE}

The left side is opposite over hypotenuse in the first triangle; the right side is opposite over hypotenuse in the second. They are equal. So "opposite over hypotenuse for a

40^\circ

angle" is a single number, no matter how big the triangle is. The same argument works for cosine and tangent, because in similar figures the ratio of any two sides of one triangle equals the ratio of the corresponding two sides of the other.

This is why

\sin

\cos

, and

\tan

can be treated as functions of the angle: the input is an angle measure, the output is a ratio, and the triangle's size is irrelevant.

A useful consequence: the ratios are pure numbers with no units. Centimeters divided by centimeters cancel. If a classmate reports that

\cos 60^\circ=0.5

inches, something has gone wrong.

Another consequence worth remembering: since the hypotenuse is the longest side, both

\sin\theta

and

\cos\theta

for an acute angle must be less than 1. Tangent has no such ceiling — it can be any positive number, and it grows without bound as the angle approaches

90^\circ

Computing Ratios from a Triangle

When a problem hands you side lengths, finding the three ratios is mostly bookkeeping — but do it carefully.

Start by making sure you have all three sides. If only two are given, use the Pythagorean Theorem to get the third before writing ratios that need it. Then choose your angle, label the sides relative to that angle, and write each fraction in lowest terms.

Given	What to write for angle $A$
leg opposite $A$ = 6, hypotenuse = 10	$\sin A=\frac{6}{10}=\frac{3}{5}$
leg adjacent to $A$ = 8, hypotenuse = 10	$\cos A=\frac{8}{10}=\frac{4}{5}$
legs 6 and 8, with 6 opposite $A$	$\tan A=\frac{6}{8}=\frac{3}{4}$

Leave answers as exact fractions or exact radicals unless the problem asks for a decimal. If a side is

5\sqrt{2}

and the hypotenuse is

10

, then

\sin A=\frac{5\sqrt{2}}{10}=\frac{\sqrt{2}}{2}

. Rounding too early is a common source of wrong final answers in later multi-step problems.

A quick self-check: the ratio for the larger acute angle should have the larger sine, because the larger angle faces the longer leg. If your triangle has a clearly larger angle at

A

but you computed

\sin A<\sin B

, you swapped a label.

When you use a calculator to evaluate something like

\tan 37^\circ

, make sure the calculator is in degree mode. You should see

0.7536

; if instead you see

-0.8408

, the calculator is in radians and every answer in the assignment will be wrong.

Sine and Cosine of Complementary Angles

In any right triangle, the two acute angles add to

90^\circ

— they are complementary, because the three angles total

180^\circ

and one of them is the right angle. Label them

A

and

B

, so

B=90^\circ-A

.

Now look at what the legs are called from each vantage point. The leg opposite

A

is the leg adjacent to

B

, and the leg adjacent to

A

is the leg opposite

B

. The hypotenuse is shared. Therefore

\sin A=\frac{a}{c}=\cos B \qquad\text{and}\qquad \cos A=\frac{b}{c}=\sin B

Since

B=90^\circ-A

, this is usually written as the cofunction identity:

\sin\theta=\cos(90^\circ-\theta)\qquad \cos\theta=\sin(90^\circ-\theta)

That is literally where the name "cosine" comes from: the sine of the complement.

So

\sin 25^\circ=\cos 65^\circ

exactly — not approximately, and not because of any calculator rounding. Check it: both are about

0.4226

.

This identity shows up in equation problems. If

\sin(3x+10)^\circ=\cos(2x)^\circ

, the two angles must be complementary, so

(3x+10)+2x=90

, giving

5x=80

and

x=16

.

Two cautions. First, the identity pairs sine with cosine only — it is not true that

\tan\theta=\tan(90^\circ-\theta)

. In fact

\tan(90^\circ-\theta)=\frac{1}{\tan\theta}

, since the opposite and adjacent legs trade places. Second, complementary means the measures sum to

90

, not that the two expressions are equal to each other. Writing

3x+10=2x

is a common error that produces a negative angle.

Key terms

Hypotenuse.: The side of a right triangle opposite the right angle; it is always the longest side and is never called a leg.
Opposite leg.: Relative to a chosen acute angle, the leg that does not form part of that angle.
Adjacent leg.: Relative to a chosen acute angle, the leg that forms one side of that angle (the hypotenuse does not count as adjacent).
Sine of an acute angle.: The ratio $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , a number strictly between 0 and 1 for acute angles.
Cosine of an acute angle.: The ratio $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ , also strictly between 0 and 1 for acute angles.
Tangent of an acute angle.: The ratio $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ ; it equals the slope of a line making angle $\theta$ with the horizontal and can exceed 1.
Cofunction identity.: The relationship $\sin\theta=\cos(90^\circ-\theta)$ and $\cos\theta=\sin(90^\circ-\theta)$ , which follows from the two acute angles of a right triangle being complementary.
AA Similarity.: If two angles of one triangle are congruent to two angles of another, the triangles are similar; this is why trigonometric ratios depend only on the angle.

Worked example

In right triangle

PQR

, the right angle is at

Q

PQ=8

, and

QR=15

. Find

\sin P

\cos P

, and

\tan P

as exact fractions, then find

\sin R

and

\cos R

and explain how they relate to the ratios for angle

P

Step 1: Find the missing side. The right angle is at

Q

, so the hypotenuse is

PR

. By the Pythagorean Theorem,

PR^2=8^2+15^2=64+225=289

, so

PR=17

.

Step 2: Label the sides from angle

P

. Angle

P

is formed by sides

PQ

and

PR

. The hypotenuse is

PR=17

. The leg touching

P

PQ=8

, so that is adjacent. The remaining leg,

QR=15

, is opposite

P

.

Step 3: Write the ratios for

P

\sin P=\frac{15}{17},\qquad \cos P=\frac{8}{17},\qquad \tan P=\frac{15}{8}

None of these reduce, since 8, 15, and 17 share no common factors.

Step 4: Re-label from angle

R

. This is the step students skip. Angle

R

is formed by

QR

and

PR

. The hypotenuse is still

PR=17

. Now

QR=15

is adjacent to

R

, and

PQ=8

is opposite

R

.

Step 5: Write the ratios for

R

\sin R=\frac{8}{17},\qquad \cos R=\frac{15}{17}

Step 6: Interpret. Notice

\sin P=\cos R=\frac{15}{17}

and

\cos P=\sin R=\frac{8}{17}

. Angles

P

and

R

are complementary because

P+R=180^\circ-90^\circ=90^\circ

, so the cofunction identity

\sin P=\cos(90^\circ-P)=\cos R

holds exactly. A quick sanity check:

\tan P=\frac{15}{8}>1

because the leg opposite

P

is longer than the leg adjacent to it, meaning

P

is the larger acute angle.

Practice questions

In right triangle

ABC

with the right angle at

C

\sin A=\frac{7}{25}

. What is

\cos B

$\frac{7}{25}$
$\frac{24}{25}$
$\frac{25}{7}$
$\frac{7}{24}$

Answer: $\frac{7}{25}$

Angles

A

and

B

are the two acute angles of a right triangle, so

A+B=90^\circ

and

B=90^\circ-A

. The cofunction identity gives

\cos B=\cos(90^\circ-A)=\sin A=\frac{7}{25}

. You can also see it directly from the sides: the leg opposite

A

is the leg adjacent to

B

, and the hypotenuse is shared, so the two ratios are the same fraction. The value

\frac{24}{25}

is what you get if you compute the third side and then find

\sin B

instead of

\cos B

— a common slip when the labels are not redrawn for the new angle.

Solve for

x

\cos(4x+6)^\circ=\sin(x-1)^\circ

, where both angles are acute. Then state the measure of each angle.

Answer: $x=17$ , so the angles measure $74^\circ$ and $16^\circ$ .

Cosine of one angle equals sine of another exactly when the two angles are complementary, so set the sum equal to 90:

(4x+6)+(x-1)=90

. Combine like terms to get

5x+5=90

, so

5x=85

and

x=17

. Substituting back,

4(17)+6=74

and

17-1=16

. Check:

74+16=90

, so the angles are complementary and

\cos 74^\circ=\sin 16^\circ\approx 0.2756

. The most common error is writing

4x+6=x-1

, which treats the identity as if it said the angles are equal; that gives

x=-\frac{7}{3}

and produces negative angle measures, an immediate signal that the setup was wrong.

Two right triangles each contain a

32^\circ

angle. The first has a hypotenuse of 5 units; the second has a hypotenuse of 40 units. Explain why

\sin 32^\circ

is the same number in both triangles, and state what that means about the leg opposite the

32^\circ

angle in each.

Answer: By AA Similarity the triangles are similar, so corresponding sides are proportional and opposite-over-hypotenuse is identical; the opposite legs are $5\sin 32^\circ$ and $40\sin 32^\circ$ , so the second is exactly 8 times the first.

Each triangle has a right angle and a

32^\circ

angle, which is two pairs of congruent angles, so AA Similarity makes them similar. In similar triangles corresponding sides are proportional, meaning

\frac{\text{opp}_1}{\text{hyp}_1}=\frac{\text{opp}_2}{\text{hyp}_2}

. That common value is what we call

\sin 32^\circ

, roughly

0.5299

. Multiplying gives opposite legs of about

2.65

and about

21.19

units. The scale factor from the first triangle to the second is

\frac{40}{5}=8

, and indeed

21.19\approx 8(2.65)

. This is the whole justification for using a single sine value for every triangle with a given acute angle.

FAQ

Why do sine and cosine of an acute angle always come out less than 1?: Both ratios have the hypotenuse in the denominator, and the hypotenuse is the longest side of a right triangle. A fraction whose numerator is a leg and whose denominator is the hypotenuse must be less than 1. If you ever compute $\sin\theta=\frac{13}{5}$ for an acute angle, you have flipped a fraction or mislabeled the hypotenuse. Tangent is different: it compares two legs, so it can be less than 1, equal to 1 (at $45^\circ$ ), or arbitrarily large.
Does the size of the triangle change the value of a trigonometric ratio?: No. Any two right triangles sharing an acute angle are similar by AA, so their corresponding sides are proportional and every side-to-side ratio matches. Doubling the triangle doubles both the numerator and the denominator, leaving the fraction unchanged. This is exactly why a calculator can report a single value for $\cos 41^\circ$ without asking you how big your triangle is.
What is the difference between $\sin\theta$ and $\sin^{-1}$ on my calculator?: The $\sin$ key takes an angle in and returns a ratio. The $\sin^{-1}$ key (inverse sine) takes a ratio in and returns the angle. In this lesson you mostly go in the first direction — angle to ratio — and set up ratios from given side lengths. Finding the angle from a ratio is the focus of the next lesson on solving right triangles. Note that $\sin^{-1}$ does not mean $\frac{1}{\sin}$ .
Is there a cofunction rule for tangent like the one for sine and cosine?: Yes, but it is a reciprocal rather than an equality. Since swapping to the complementary angle trades the opposite and adjacent legs, $\tan(90^\circ-\theta)=\frac{\text{adjacent}}{\text{opposite}}=\frac{1}{\tan\theta}$ . That reciprocal is called cotangent. So $\tan 30^\circ$ and $\tan 60^\circ$ are not equal — their product is 1.

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