In short

For a right triangle with an acute angle \theta, the six trigonometric ratios — sine, cosine, tangent, and their reciprocals — are specific ratios of the triangle's sides. They depend only on the angle, not on the size of the triangle. The identity \sin^2\theta + \cos^2\theta = 1 follows directly from the Pythagorean theorem, and the exact values at , 30°, 45°, 60°, and 90° can all be derived from two triangles you can draw by hand.

A ladder leans against a wall. The ladder is 5 metres long and its foot is 3 metres from the base of the wall. How high up the wall does the ladder reach?

By the Pythagorean theorem: height = \sqrt{25 - 9} = 4 metres. You have known how to do this since class 8.

Now change the question. The ladder is still 5 metres long, but instead of being told the foot-to-wall distance, you are told the angle the ladder makes with the ground: 53°.

The same right triangle, but the given information has shifted from a side to an angle. To convert the angle back into a side length — to turn 53° into a number of metres — you need a bridge between angles and lengths. That bridge is a trigonometric ratio.

The idea is ancient. Indian astronomers needed it to compute the positions of planets and the lengths of shadows. Aryabhata, writing in 499 CE, tabulated a quantity he called ardha-jya (half-chord) — which is exactly what we now call the sine. The name changed as the idea moved west through Arabic and Latin translations, but the mathematics is the same.

The six ratios — defined from a right triangle

Take a right triangle. Label one of the acute angles \theta. The three sides relative to this angle are:

A right triangle with sides labelled relative to angle thetaA right triangle with the right angle at the bottom-right. The angle theta is at the bottom-left. The bottom side is labelled Adjacent, the right side is labelled Opposite, and the slanted side from bottom-left to top-right is labelled Hypotenuse. A small square marks the right angle. θ Adjacent Opposite Hypotenuse
A right triangle with its sides named relative to the angle $\theta$. The hypotenuse is always the longest side. "Opposite" and "Adjacent" swap if you look at the other acute angle.

The six trigonometric ratios are defined as:

The six trigonometric ratios

For an acute angle \theta in a right triangle:

\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}
\csc\theta = \frac{\text{Hypotenuse}}{\text{Opposite}}, \quad \sec\theta = \frac{\text{Hypotenuse}}{\text{Adjacent}}, \quad \cot\theta = \frac{\text{Adjacent}}{\text{Opposite}}

The first three are the primary ratios. The last three are their reciprocals: \csc\theta = 1/\sin\theta, \sec\theta = 1/\cos\theta, \cot\theta = 1/\tan\theta.

The mnemonic that generations of Indian students have used: SOH-CAH-TOA — Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

Notice two things. First, these ratios are numbers — pure numbers, with no units. They are the ratio of two lengths, so the metres cancel. Second, the ratios depend only on the angle \theta, not on the size of the triangle. If you scale the triangle up (longer ladder, farther wall), all three sides scale by the same factor, and the ratios stay the same. This is the key insight: similar triangles have the same trigonometric ratios. The angle determines the shape; the shape determines the ratios.

Why the ratios depend only on the angle

This deserves a careful argument, because it is the foundation of all trigonometry.

Draw two right triangles that share the same acute angle \theta but have different sizes. Call the sides of the smaller triangle a, b, c (with c the hypotenuse) and the sides of the larger triangle ka, kb, kc for some scaling factor k > 1. These two triangles are similar — they have the same angles, and their sides are proportional.

Compute \sin\theta for the larger triangle:

\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{ka}{kc} = \frac{a}{c}

The k cancels. You get the same value as for the smaller triangle. The same argument works for all six ratios.

This means \sin 30° is a single, definite number — not "it depends on the triangle." Every right triangle with a 30° angle, from a tiny one drawn on a postage stamp to one whose hypotenuse stretches across a cricket pitch, has \sin 30° = 1/2. The ratio is a property of the angle itself.

The reciprocal ratios

The last three ratios — cosecant, secant, and cotangent — are simply the reciprocals of the first three:

\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}

And there is one more relationship that connects sine, cosine, and tangent:

\tan\theta = \frac{\sin\theta}{\cos\theta}

This follows directly from the definitions: \dfrac{\sin\theta}{\cos\theta} = \dfrac{O/H}{A/H} = \dfrac{O}{A} = \tan\theta. Similarly, \cot\theta = \dfrac{\cos\theta}{\sin\theta}.

These are not new identities — they are just the definitions rewritten. But they are used so constantly that they are worth committing to memory.

The Pythagorean identities — proved

Here is where the definitions connect to the most fundamental theorem in geometry.

The first identity

Take a right triangle with sides Opposite = a, Adjacent = b, Hypotenuse = c. By the Pythagorean theorem:

a^2 + b^2 = c^2

Divide both sides by c^2:

\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1
\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1

But a/c = \sin\theta and b/c = \cos\theta. So:

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

That is the entire proof — three lines. The Pythagorean theorem becomes a statement about sine and cosine. This identity holds for every angle \theta (not just acute angles — it extends to all angles once you define sine and cosine using the unit circle, which you will see in Trigonometric Ratios of Any Angle).

The second identity

Divide \sin^2\theta + \cos^2\theta = 1 by \cos^2\theta (assuming \cos\theta \neq 0):

\frac{\sin^2\theta}{\cos^2\theta} + 1 = \frac{1}{\cos^2\theta}
\tan^2\theta + 1 = \sec^2\theta

Or equivalently: \sec^2\theta - \tan^2\theta = 1.

The third identity

Divide \sin^2\theta + \cos^2\theta = 1 by \sin^2\theta (assuming \sin\theta \neq 0):

1 + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}
1 + \cot^2\theta = \csc^2\theta

Or equivalently: \csc^2\theta - \cot^2\theta = 1.

The three Pythagorean identities

\sin^2\theta + \cos^2\theta = 1
1 + \tan^2\theta = \sec^2\theta
1 + \cot^2\theta = \csc^2\theta

All three are the Pythagorean theorem in disguise. The first is fundamental; the other two are obtained by dividing through by \cos^2\theta and \sin^2\theta respectively.

These identities are not facts to memorise as three separate items. There is one identity (\sin^2 + \cos^2 = 1), and the other two are automatic consequences. If you forget the second or third, re-derive them in five seconds by dividing.

Values for standard angles

Five angles come up so often that their trigonometric ratios are worth knowing exactly: , 30°, 45°, 60°, and 90°. All five can be derived from two triangles.

The 45-45-90 triangle

Take a square with side length 1. Cut it along the diagonal. You get a right triangle with two 45° angles, legs of length 1, and hypotenuse \sqrt{1^2 + 1^2} = \sqrt{2}.

The 45-45-90 triangle from a unit squareA right isosceles triangle with legs of length 1 and hypotenuse of length root 2. The two acute angles are both 45 degrees. A dashed square outline shows how the triangle is half of a unit square. 1 1 √2 45° 45°
Cut a unit square along its diagonal and you get the 45-45-90 triangle: legs $1$, hypotenuse $\sqrt{2}$.

Read off the ratios for 45°:

\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \quad \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \quad \tan 45° = \frac{1}{1} = 1

Sine and cosine are equal at 45° — because the triangle is isosceles, and the opposite and adjacent sides are the same.

The 30-60-90 triangle

Take an equilateral triangle with side length 2. Drop a perpendicular from one vertex to the opposite side. This cuts the equilateral triangle into two congruent right triangles, each with angles 30°, 60°, 90°.

The hypotenuse of each right triangle is a side of the equilateral triangle: 2. The short leg (the half-base) is 1. The long leg (the altitude) is \sqrt{4 - 1} = \sqrt{3}.

The 30-60-90 triangle from an equilateral triangleAn equilateral triangle with side 2, bisected by a vertical altitude. The left half is a 30-60-90 triangle with hypotenuse 2, short leg 1, and long leg root 3. The 30-degree angle is at the top, the 60-degree angle is at the bottom-left. 1 2 √3 30° 60°
Bisect an equilateral triangle of side $2$ and you get the 30-60-90 triangle: legs $1$ and $\sqrt{3}$, hypotenuse $2$.

Read off the ratios for 30° (opposite side = 1, adjacent side = \sqrt{3}, hypotenuse = 2):

\sin 30° = \frac{1}{2}, \quad \cos 30° = \frac{\sqrt{3}}{2}, \quad \tan 30° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Read off the ratios for 60° (opposite side = \sqrt{3}, adjacent side = 1, hypotenuse = 2):

\sin 60° = \frac{\sqrt{3}}{2}, \quad \cos 60° = \frac{1}{2}, \quad \tan 60° = \sqrt{3}

Notice: \sin 30° = \cos 60° and \sin 60° = \cos 30°. This is not a coincidence — in a right triangle, the two acute angles add up to 90°, and the side that is "opposite" to one angle is "adjacent" to the other. In general, \sin\theta = \cos(90° - \theta) for any acute angle \theta. That is why cosine is called co-sine — it is the sine of the complement.

The boundary cases: 0° and 90°

These cannot come from a physical right triangle (a triangle with a angle is a line segment, not a triangle), but they can be defined as limits.

As \theta \to 0°: the opposite side shrinks to zero and the adjacent side approaches the hypotenuse.

\sin 0° = 0, \quad \cos 0° = 1, \quad \tan 0° = 0

As \theta \to 90°: the adjacent side shrinks to zero and the opposite side approaches the hypotenuse.

\sin 90° = 1, \quad \cos 90° = 0, \quad \tan 90° = \text{undefined}

Tangent of 90° is undefined because you would be dividing by zero (\cos 90° = 0). This is not a flaw — it reflects a geometric fact. When \theta = 90°, the "triangle" has collapsed: the opposite side and the hypotenuse are the same line, and the adjacent side has vanished. A ratio involving a vanished side is not a real number.

The complete table

\theta \sin\theta \cos\theta \tan\theta \csc\theta \sec\theta \cot\theta
0 1 0 undefined 1 undefined
30° \dfrac{1}{2} \dfrac{\sqrt{3}}{2} \dfrac{1}{\sqrt{3}} 2 \dfrac{2}{\sqrt{3}} \sqrt{3}
45° \dfrac{1}{\sqrt{2}} \dfrac{1}{\sqrt{2}} 1 \sqrt{2} \sqrt{2} 1
60° \dfrac{\sqrt{3}}{2} \dfrac{1}{2} \sqrt{3} \dfrac{2}{\sqrt{3}} 2 \dfrac{1}{\sqrt{3}}
90° 1 0 undefined 1 undefined 0

Do not memorise this table as 30 separate entries. Instead, remember two things:

  1. The sine values for through 90° are \dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2} — that is, 0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1.

  2. The cosine values are the same list in reverse order (because \cos\theta = \sin(90° - \theta)).

Everything else follows from \tan = \sin/\cos and the reciprocal relations.

Worked examples

Example 1: Finding a missing side using trigonometric ratios

A flagpole casts a shadow 10 metres long when the sun makes an angle of 60° with the ground. Find the height of the flagpole.

Step 1. Draw the right triangle. The flagpole is the vertical side (opposite the 60° angle). The shadow is the horizontal side (adjacent to the 60° angle). The sun ray is the hypotenuse.

Why: identifying which side is opposite and which is adjacent to the given angle determines which ratio to use.

Step 2. Choose the right ratio. You know the adjacent side (10 m) and want the opposite side. The ratio connecting opposite and adjacent is tangent.

\tan 60° = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{10}

Why: tangent is the only ratio that involves both opposite and adjacent without involving the hypotenuse, which you do not know.

Step 3. Substitute the known value of \tan 60°.

\sqrt{3} = \frac{h}{10}
h = 10\sqrt{3} \approx 17.32 \text{ m}

Why: \tan 60° = \sqrt{3} from the 30-60-90 triangle derived earlier.

Step 4. Verify using the Pythagorean theorem. The hypotenuse (sun ray distance) would be \sqrt{100 + 300} = \sqrt{400} = 20 m. Check: \sin 60° = \dfrac{10\sqrt{3}}{20} = \dfrac{\sqrt{3}}{2}. Correct.

Result: The flagpole is 10\sqrt{3} \approx 17.32 metres tall.

The flagpole (red, vertical) is $10\sqrt{3}$ metres tall. The shadow (horizontal) is $10$ metres. The sun's ray makes a $60°$ angle with the ground.

The height is \sqrt{3} times the shadow length — exactly the tangent of 60°. If the sun angle were 45° instead, the height and shadow would be equal (since \tan 45° = 1). A steeper sun angle means a shorter shadow for the same pole.

Example 2: Using Pythagorean identities to find all ratios from one

Given that \sin\theta = \dfrac{3}{5} and \theta is an acute angle, find all six trigonometric ratios.

Step 1. Use the Pythagorean identity to find \cos\theta.

\sin^2\theta + \cos^2\theta = 1
\frac{9}{25} + \cos^2\theta = 1
\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}
\cos\theta = \frac{4}{5}

Why: the positive root is taken because \theta is acute, so \cos\theta > 0.

Step 2. Compute \tan\theta.

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}

Why: \tan = \sin/\cos is faster than going back to the triangle.

Step 3. Compute the reciprocal ratios.

\csc\theta = \frac{1}{\sin\theta} = \frac{5}{3}, \quad \sec\theta = \frac{1}{\cos\theta} = \frac{5}{4}, \quad \cot\theta = \frac{1}{\tan\theta} = \frac{4}{3}

Why: reciprocal ratios are just flipped fractions — no new computation needed.

Step 4. Verify with the other Pythagorean identities.

1 + \tan^2\theta = 1 + \dfrac{9}{16} = \dfrac{25}{16} = \sec^2\theta. \checkmark

1 + \cot^2\theta = 1 + \dfrac{16}{9} = \dfrac{25}{9} = \csc^2\theta. \checkmark

Result: The six ratios are \sin\theta = 3/5, \cos\theta = 4/5, \tan\theta = 3/4, \csc\theta = 5/3, \sec\theta = 5/4, \cot\theta = 4/3.

The 3-4-5 right triangle. With $\sin\theta = 3/5$, the opposite side is $3$, the adjacent side is $4$, and the hypotenuse is $5$. All six ratios are readable from this one triangle.

The 3-4-5 triangle is the most famous Pythagorean triple — and it shows that once you know one trigonometric ratio and the quadrant, you know all six. The Pythagorean identity is the tool that makes the jump from one ratio to the rest.

Common confusions

Going deeper

If you came here to learn the six ratios, the identities, and the standard angle values, you have everything you need for class 10 and the first stage of trigonometry. What follows is for readers who want to see the bigger picture.

Why these particular five angles?

The angles 30°, 45°, 60° are special because their trigonometric ratios are expressible in terms of square roots — they are algebraic numbers. Most angles do not have this property. Try computing \sin 1° exactly — you cannot write it as a finite expression using +, -, \times, \div, and \sqrt{\ }. (It is an algebraic number, but writing it out requires nested radicals that fill a page.)

The angles that do have clean exact values are connected to regular polygons that can be constructed with ruler and compass. A regular hexagon gives you 30° and 60°. A square gives you 45°. A regular pentagon gives you 36° and 72°, whose sines and cosines involve the golden ratio \phi = (1 + \sqrt{5})/2. The 17-gon, the 257-gon, and the 65537-gon also yield constructible angles — Gauss proved this in 1796 when he was 19.

Aryabhata's sine table

Aryabhata's Aryabhatiya (499 CE) contains a table of sine values for angles from to 90° in steps of 3.75° (that is, 3°45'). He called the quantity ardha-jya (half-chord), which is geometrically exact: if you draw a chord in a circle, the sine of the angle is half the chord length divided by the radius.

The name's journey is a story about translation. Arabic mathematicians transliterated jya (the Sanskrit for chord) as jiba. Scribes later misread this as jaib (Arabic for "bay" or "fold"), and when European translators turned it into Latin, they used sinus (Latin for "bay" or "curve"). From sinus came the English sine. The original Sanskrit meaning — half-chord — is hidden under layers of mistranslation, but the mathematics has been exact from the beginning.

Bhaskara II, in his Lilavati (1150 CE), systematised the computation of sines and cosines and stated the relationship \sin^2\theta + \cos^2\theta = R^2 (where R is the radius), which is the Pythagorean identity scaled by R^2.

The unit circle — a preview

The right-triangle definition has a limitation: it only works for acute angles (0° < \theta < 90°). To handle angles like 150° or -45° or 720°, you need a new definition that agrees with the triangle definition for acute angles but extends beyond.

The answer is the unit circle — a circle of radius 1 centred at the origin. Place a point on the circle at angle \theta measured counterclockwise from the positive x-axis. The x-coordinate of that point is \cos\theta, and the y-coordinate is \sin\theta.

For acute angles, this reproduces exactly the right-triangle definition (the right triangle sits inside the first quadrant of the unit circle). For obtuse angles, the x-coordinate becomes negative, giving \cos\theta < 0. For angles beyond 360°, the point wraps around the circle again, making sine and cosine periodic. The full story is in Trigonometric Ratios of Any Angle.

The unit circle with the five standard angles marked. For the $30°$ point, the vertical dashed line is $\sin 30° = 1/2$ and the horizontal dashed line is $\cos 30° = \sqrt{3}/2$. Every point on the circle has coordinates $(\cos\theta, \sin\theta)$.

From ratios to functions

Once you have the unit-circle definition, \sin and \cos become functions — they accept any real number as input and return a value between -1 and 1. Their graphs are the famous sine and cosine waves: smooth, periodic curves that repeat every 360° (or 2\pi radians). These waves model everything from sound and light to alternating current and the motion of a pendulum. The article Trigonometric Functions and Graphs explores this.

Where this leads next

You now know the six trigonometric ratios, their identities, and their values at the standard angles. The next steps: