Drag point P round the circle, or drag along the graph to pick an angle
Controls
Angle unit
Readings
- Angle, θ
- 30.0°
- θ in radians
- π/6
- sin θ (height)
- 0.5000
- cos θ (across)
- 0.8660
- tan θ
- 0.5774
- Quadrant
- Quadrant I
- Reference angle
- 30.0°
- sin²θ + cos²θ
- 1.0000
- Exact value (sin, cos)
- 1/2, √3/2
How to use this simulation
- Drag the purple point P round the circle. The green segment is cos θ, the red one is sin θ and the amber one is tan θ.
- You can also drag along the graph on the right: left to right runs from 0° to 360°.
- Turn on "snap to standard angles" and stop at 30°, 45° and 60° to see exact values such as √3/2 and 1/√2.
- Press ▶ and P turns at a steady rate while the sin and cos waves are drawn; slow it to 0.25× to follow it.
- Switch the angle unit to radians and the graph axis reads π/2, π, 3π/2 and 2π.
- Watch the sin²θ + cos²θ reading: it stays at 1 for every angle, which is the key identity.
Ferris wheels, shadows and tree heights: where trigonometry lives
Ever ridden a Ferris wheel? You travel round a circle, and your height keeps changing: up, over the top, down, and up again. How far above the centre you are is the sine of your angle; how far left or right of the centre you are is the cosine. The simulation on this page is exactly that Ferris wheel, just with a radius of 1.
Another one: measure a pole’s shadow and the angle the sun makes with the ground, and you can work out how tall the pole is without climbing it. People in ancient Egypt measured the height of a pyramid this way. Where a tape measure cannot reach, trigonometry can.
The word comes from Greek for "triangle measuring". But you will soon see it is not just about triangles: spinning wheels, waves, sound, alternating current and computer graphics all speak the language of sin and cos.
Trigonometric ratios in a right triangle
Take a right triangle and pick one of its acute angles, θ. Now name the sides from θ’s point of view: the side across from θ is the opposite, the side next to θ that is not the longest is the adjacent, and the longest side, across from the right angle, is the hypotenuse.
Notice that opposite and adjacent depend on which angle you picked. Switch to the other acute angle and they swap places, while the hypotenuse stays the same. This swap is one of the most common exam mistakes.
Take the sides two at a time and you get six ratios: sine, cosine and tangent, and their flips cosecant, secant and cotangent. The best part: a bigger or smaller triangle with the same angle gives the same ratios, because similar triangles scale every side by the same factor. So a ratio depends only on the angle.
sin θ = opposite / hypotenusesine
cos θ = adjacent / hypotenusecosine
tan θ = opposite / adjacent = sin θ / cos θtangent
From the triangle to the unit circle: why P = (cos θ, sin θ)
Now a clever move. Draw a circle centred at the origin O with radius 1: the unit circle. Start on the positive x-axis, turn anticlockwise through an angle θ, and mark the point P where you land on the circle.
Drop a perpendicular from P to the x-axis. You have a right triangle whose hypotenuse OP is 1. So sin θ = opposite/1 = the height of P, and cos θ = adjacent/1 = how far across P is. The coordinates of P are exactly (cos θ, sin θ). With a hypotenuse of 1, there is nothing to divide.
The big payoff: θ no longer has to be acute. When P moves into the second, third or fourth quadrant it still has coordinates, so 120°, 210° and even 330° all have a sine and a cosine. When a coordinate turns negative, so does the ratio.
The simulation starts at θ = 30°. There P is 0.50 high, so sin 30° = 0.50, and 0.866 across, so cos 30° ≈ 0.866. Their ratio is tan 30° ≈ 0.577. Open the readings panel and check.
Tangent has a picture too. Imagine the vertical line that touches the circle at x = 1 (a tangent line). Extend OP until it meets that line; the height of the meeting point is tan θ, which is where the name comes from. Near 90°, OP is almost vertical and never meets the line, which is why tan 90° is undefined.
Sin, cos and tan table for standard angles
0°, 30°, 45°, 60° and 90° are the standard angles, and most exam questions use them. Every value in the table below is computed in code and shown first as an exact surd, then as a decimal.
The 30° and 60° values come from an equilateral triangle. Take sides of 2 and cut it down the middle: you get two 30°-60°-90° triangles with sides 1, √3 and 2, so sin 30° = 1/2 and sin 60° = √3/2. The 45° values come from an isosceles right triangle with sides 1, 1 and √2, so sin 45° = cos 45° = 1/√2.
| Angle θ | Radians | sin θ | cos θ | tan θ |
|---|---|---|---|---|
| 0° | 0 | 0 ≈ 0.000 | 1 ≈ 1.000 | 0 ≈ 0.000 |
| 30° | π/6 | 1/2 ≈ 0.500 | √3/2 ≈ 0.866 | 1/√3 ≈ 0.577 |
| 45° | π/4 | 1/√2 ≈ 0.707 | 1/√2 ≈ 0.707 | 1 ≈ 1.000 |
| 60° | π/3 | √3/2 ≈ 0.866 | 1/2 ≈ 0.500 | √3 ≈ 1.732 |
| 90° | π/2 | 1 ≈ 1.000 | 0 ≈ 0.000 | Undefined |
The four quadrants and the ASTC sign rule
The axes split the plane into four quadrants. In quadrant I (0° to 90°), x and y are both positive, so every ratio is positive. In quadrant II (90° to 180°), x is negative and y positive, so only sine (and cosecant) is positive. In quadrant III (180° to 270°), both are negative, so their quotient, tangent, is positive. In quadrant IV (270° to 360°), only x is positive, so cosine is.
The memory line is ASTC: All Students Take Calculus (some teachers say All Silver Tea Cups). In the simulation, the quadrant holding P lights up and shows which ratio is positive there.
To find any angle’s ratio, use three steps: (1) find the quadrant, (2) find the reference angle, the acute angle the terminal side makes with the x-axis, and (3) write that acute angle’s value and give it the quadrant’s sign. The table below is computed exactly that way.
| Angle θ | Quadrant | Reference angle | sin θ | cos θ | tan θ |
|---|---|---|---|---|---|
| 120° | Quadrant II | 60° | √3/2 ≈ 0.866 | −1/2 ≈ -0.500 | −√3 ≈ -1.732 |
| 135° | Quadrant II | 45° | 1/√2 ≈ 0.707 | −1/√2 ≈ -0.707 | −1 ≈ -1.000 |
| 150° | Quadrant II | 30° | 1/2 ≈ 0.500 | −√3/2 ≈ -0.866 | −1/√3 ≈ -0.577 |
| 210° | Quadrant III | 30° | −1/2 ≈ -0.500 | −√3/2 ≈ -0.866 | 1/√3 ≈ 0.577 |
| 240° | Quadrant III | 60° | −√3/2 ≈ -0.866 | −1/2 ≈ -0.500 | √3 ≈ 1.732 |
| 300° | Quadrant IV | 60° | −√3/2 ≈ -0.866 | 1/2 ≈ 0.500 | −√3 ≈ -1.732 |
| 330° | Quadrant IV | 30° | −1/2 ≈ -0.500 | √3/2 ≈ 0.866 | −1/√3 ≈ -0.577 |
Reciprocal ratios: cosec, sec and cot
Flip the first three ratios and you get the other three. Cosecant is the reciprocal of sine, secant of cosine, and cotangent of tangent. A handy pairing rule: each pair has exactly one "co" between them, so sin goes with cosec and cos goes with sec.
When a ratio is zero, its reciprocal is undefined, because you cannot divide by zero. sin 0° = 0, so cosec 0° is undefined; cos 90° = 0, so sec 90° and tan 90° are undefined.
cosec θ = 1 / sin θ = hypotenuse / opposite
sec θ = 1 / cos θ = hypotenuse / adjacent
cot θ = 1 / tan θ = cos θ / sin θ
Trigonometric identities and formulas
On the unit circle, P = (cos θ, sin θ) is 1 away from the centre. Apply Pythagoras and you get cos²θ + sin²θ = 1. This is the single most important identity in trigonometry, and the readings panel shows it equal to 1 at every angle.
Divide both sides by cos²θ and you get 1 + tan²θ = sec²θ. Divide by sin²θ instead and you get 1 + cot²θ = cosec²θ. The three Pythagorean identities are really one identity in three outfits.
The complementary-angle formulas matter too. In a right triangle the two acute angles add to 90°, and one angle’s opposite is the other’s adjacent. So sin(90° − θ) = cos θ and cos(90° − θ) = sin θ, which is why sin 30° = cos 60° = 1/2.
sin²θ + cos²θ = 1the Pythagorean identity
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ
sin(90° − θ) = cos θ, cos(90° − θ) = sin θcomplementary angles
sin(180° − θ) = sin θ, cos(180° − θ) = −cos θallied angles
Degrees and radians
In degrees, a full turn is 360°. Radians measure differently: one radian is the angle at the centre when the arc you travel along is as long as the radius. A full circle’s circumference is 2πr, so a full turn is 2π radians.
So π radians = 180°. To go from degrees to radians multiply by π/180; to go back multiply by 180/π. One radian ≈ 57.30°, just under 57 degrees.
On the unit circle radians have a lovely meaning: with a radius of 1, the length of arc P has travelled is the angle in radians. Higher maths and calculus use radians almost all the time, so practise with the unit switch in the simulation.
π rad = 180°
radians = degrees × π / 180
degrees = radians × 180 / π
Graphs of sin, cos and tan
As P turns, its height (sin θ) rises from 0 to 1, falls back to 0, goes down to −1 and back to 0. Plot that against the angle and you get a smooth wave: the sine graph. In the simulation a dashed line runs straight across from P to the graph, because the circle’s height is the graph’s height.
The cosine graph is the same wave shifted by 90°: it starts at cos 0° = 1 and falls. Both waves have an amplitude of 1 and a period of 360° (2π), meaning the pattern repeats every full turn.
The tangent graph is different. It runs from −∞ to +∞ and breaks at 90° and 270°, where cos θ = 0, and its period is only 180°. Turn on the tangent option and watch the amber curve.
- sin θ: range −1 to 1, period 360°, starts at 0
- cos θ: range −1 to 1, period 360°, starts at 1
- tan θ: range all real numbers, period 180°, undefined at 90° and 270°
Try this in the simulation
You remember what you do far better than what you read. Try these five experiments and write down what you see.
- Put P at 45°: the red and green segments are equal, because sin 45° = cos 45°.
- Compare the red segment at 30° and at 150°: equal, because sin(180° − θ) = sin θ.
- Bring P very close to 90°: the amber tangent segment shoots up, and at exactly 90° the reading says "undefined".
- Stop at any angle and check sin²θ + cos²θ: always 1.
- Press ▶ and let it make a full turn: the sin and cos waves run 90° out of step with each other.
Solved examples
Every answer below is computed in code. Try each one yourself first, then check.
Example 1: height of a tower
Standing 20 m from the foot of a tower, you see its top at an angle of elevation of 60°. How tall is it?
Adjacent = 20 m, opposite = h, angle 60°. Opposite and adjacent means tangent: tan 60° = h/20. tan 60° ≈ 1.732, so h = 20 × 1.732 ≈ 34.64 m.
Example 2: height of a kite
A kite string is 50 m long and makes 30° with the ground. How high is the kite?
The string is the hypotenuse and the height is the opposite, so sin 30° = h/50, giving h = 50 × 1/2 = 25 m.
Example 3: all ratios from one
If sin θ = 3/5 and θ is acute, find the other five ratios.
Opposite 3, hypotenuse 5. By Pythagoras, adjacent = √(5² − 3²) = 4. So cos θ = 4/5 = 0.80, tan θ = 3/4 = 0.75, cosec θ ≈ 1.667, sec θ = 1.25 and cot θ ≈ 1.333.
Example 4: evaluate with standard values
Find sin²30° + cos²60° + tan²45°.
(1/2)² + (1/2)² + (1)² = 1/4 + 1/4 + 1 = 1.50.
Example 5: spot the identity
Find (1 − tan²30°)/(1 + tan²30°).
tan 30° = 1/√3, so tan²30° = 1/3. The value is (1 − 1/3)/(1 + 1/3) = (2/3)/(4/3) = 0.50. That equals cos 60°, because (1 − tan²θ)/(1 + tan²θ) = cos 2θ.
Example 6: allied angles
Find sin 150°, cos 210° and tan 315°.
150° is in quadrant II with reference angle 30°, and sine is positive: sin 150° = 0.50. 210° is in quadrant III with reference angle 30°, and cosine is negative: cos 210° ≈ -0.866. 315° is in quadrant IV with reference angle 45°, and tangent is negative: tan 315° = -1.00.
Example 7: degrees and radians
Convert 75° to radians and 2 radians to degrees.
75° × π/180 = 5π/12 ≈ 1.309 rad, and 2 × 180/π ≈ 114.59°.
Example 8: the sun’s angle from a shadow
A 10 m pole casts a 17.32 m shadow (that is 10√3 m). What is the sun’s angle of elevation?
tan θ = 10/(10√3) = 1/√3, so θ = 30°.
Example 9: finding the angle from its sign
If cos θ = −1/2 and θ is in quadrant II, find θ, sin θ and tan θ.
cos 60° = 1/2, so the reference angle is 60°. In quadrant II, θ = 180° − 60° = 120°. sin θ is positive ≈ 0.866 and tan θ is negative ≈ -1.732.
Example 10: a ladder against a wall
A 6 m ladder leans on a wall making 60° with the ground. How high up the wall does it reach, and how far is its foot from the wall?
Height = 6 × sin 60° = 6 × √3/2 ≈ 5.196 m. Distance of the foot = 6 × cos 60° = 3 m.
Common mistakes
Avoid these and you keep marks you would otherwise throw away.
- Mixing up opposite and adjacent: they swap when you change the angle; the hypotenuse never does.
- Reading sin²θ as sin(θ²): sin²θ means (sin θ)².
- Writing sin(A + B) = sin A + sin B: false, since sin 30° + sin 60° ≠ sin 90°.
- Leaving the calculator in radian mode for a degree question: if sin 30 is not 0.5, check the mode.
- Writing tan 90° as "infinity" or 1: the correct answer is undefined.
- Forgetting the quadrant sign: cos 120° is negative, not positive.
Trigonometry in real life
Surveyors and engineers measure angles with a theodolite to find the width of a river, the height of a hill or a boundary line. Architects set roof pitches and staircase angles with tangent.
Your phone’s GPS, a game turning a character, a photo app rotating an image: inside the code, sin and cos run thousands of times a second. Mains electricity (AC) and sound waves both follow a sine wave.
Ship and aircraft navigation, tilting solar panels to face the sun, and even analysing a cricket ball’s flight all use trigonometry.
Exam corner
In Grade 9–10 exams (SSC, GCSE, CBSE Class 10), expect three kinds of question: find the other ratios from one given ratio, prove an identity or evaluate an expression at standard angles, and a heights-and-distances problem with an angle of elevation or depression.
In Grade 11 (HSC, CBSE Class 11, A-level), the focus moves to allied angles, quadrant signs, radians and graphs, and entrance tests such as JEE and SAT ask quick questions on standard values, periods and signs.
How to prepare: rebuild the standard-angle table from scratch once a day rather than memorising it, derive the other identities from sin²θ + cos²θ = 1, and draw a diagram before every heights-and-distances problem.
Quick revision
The night before the exam, this list is enough.
- sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent
- On the unit circle P = (cos θ, sin θ); tan θ is the height on the tangent line
- sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ
- Signs: All, Sin, Tan, Cos for quadrants I to IV
- π rad = 180°; sin and cos repeat every 360°, tan every 180°
- tan 90°, sec 90°, cosec 0° and cot 0° are undefined
Frequently asked questions
What is the unit circle?
A circle of radius 1 centred at the origin. The point at angle θ on it has coordinates (cos θ, sin θ), so it shows the sine and cosine of every angle at once.
What are the six trigonometric ratios?
sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent, and their reciprocals cosec θ, sec θ and cot θ.
What are sin 30°, cos 60° and tan 45°?
sin 30° = 1/2, cos 60° = 1/2 and tan 45° = 1. For comparison, sin 45° ≈ 0.707 and sin 60° ≈ 0.866.
Why is tan 90° undefined?
tan θ = sin θ/cos θ, and cos 90° = 0. Division by zero has no value, so tan 90° is undefined.
What does ASTC mean?
It lists which ratios are positive in quadrants I to IV: All, Sin, Tan, Cos. A common way to remember it is "All Students Take Calculus".
How do I remember the sin cos tan table?
For sine, take 0, 1, 2, 3, 4, divide each by 4 and square-root it; read it backwards for cosine; divide sine by cosine for tangent.
How many degrees is one radian?
π radians = 180°, so one radian ≈ 57.30° and one degree = π/180 radians.
Why is sin²θ + cos²θ always 1?
The point (cos θ, sin θ) on the unit circle is 1 unit from the centre, so by Pythagoras cos²θ + sin²θ = 1² for every angle.
What is a reference angle?
The acute angle between the terminal side of an angle and the x-axis. Ratios of any angle equal the ratios of its reference angle, with the sign set by the quadrant.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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