Interactive guide · trigonometric ratios

Trigonometric Circle

Drag the points around the circle or type the angles. Each coloured segment is a trigonometric ratio: its (signed) length is its value. Tap any formula in the guide to see its proof.

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Angle
θ = 60°
= π/3 rad ≈1.0472 rad
Quadrant I · reference angle 60°
°
From 0° to 360°. You can also type radians using π, e.g. π/3.
Values
1 · Starting point

What the unit circle is

A circle with radius 1 centred at the origin (0, 0). An angle θ is measured from the positive x-axis, turning anticlockwise. Where the angle's terminal side crosses the circle lies the point P, and its coordinates are exactly P = (cos θ, sin θ).

DegreeA full turn is 360°.
RadianThe angle whose arc is as long as the radius. A full turn is 2π rad. To convert: rad = degrees · π / 180.See where it comes from
QuadrantEach of the 4 regions the axes split the plane into, numbered I to IV anticlockwise.
Reference angleThe acute angle between the terminal side and the x-axis. Its ratios equal those of θ, except for the sign.
2 · The six ratios

What each segment measures

Since the radius is 1, the triangle's hypotenuse is 1 and every ratio becomes a length you can see.

sin Sine

sin⁡θ=y\operatorname{sin}\theta = y

Vertical orange segment from the x-axis to P. It's the point's height. Always between −1 and 1.

See why

cos Cosine

cos⁡θ=x\operatorname{cos}\theta = x

Horizontal purple segment from the origin to P's shadow on the x-axis. Always between −1 and 1.

See why

tan Tangent

tan⁡θ=sin⁡θ/cos⁡θ\operatorname{tan}\theta = \operatorname{sin}\theta \mathbin{/} \operatorname{cos}\theta

Yellow segment on the line x = 1, which is tangent to the circle (hence the name). Undefined at 90° and 270°, because cos = 0.

See why

cot Cotangent

cot⁡θ=cos⁡θ/sin⁡θ\operatorname{cot}\theta = \operatorname{cos}\theta \mathbin{/} \operatorname{sin}\theta

Blue segment on the line y = 1. Undefined at 0° and 180°, because sin = 0.

See why

sec Secant

sec⁡θ=1/cos⁡θ\operatorname{sec}\theta = 1 \mathbin{/} \operatorname{cos}\theta

Red segment from the origin to the end of the tangent. Its absolute value is never below 1.

See why

csc Cosecant

csc⁡θ=1/sin⁡θ\operatorname{csc}\theta = 1 \mathbin{/} \operatorname{sin}\theta

Green segment from the origin to the end of the cotangent. Its absolute value is never below 1.

See why
3 · Relationships

Fundamental identities

An identity is an equality that holds for every angle (wherever the ratios exist). Under each one you'll see it checked with the current angle; tap it to see the proof.

Reciprocal

sin⁡θ⋅csc⁡θ=1\operatorname{sin}\theta \cdot \operatorname{csc}\theta = 1

Sine and cosecant are reciprocals.

0.8660 · 1.1547 = 1.0000
See proof
cos⁡θ⋅sec⁡θ=1\operatorname{cos}\theta \cdot \operatorname{sec}\theta = 1

Cosine and secant are reciprocals.

0.5000 · 2.0000 = 1.0000
See proof
tan⁡θ⋅cot⁡θ=1\operatorname{tan}\theta \cdot \operatorname{cot}\theta = 1

Tangent and cotangent are reciprocals.

1.7321 · 0.5774 = 1.0000
See proof

Quotient

tan⁡θ=sin⁡θ/cos⁡θ\operatorname{tan}\theta = \operatorname{sin}\theta \mathbin{/} \operatorname{cos}\theta

Tangent is sine divided by cosine.

0.8660 / 0.5000 = 1.7321
See proof
cot⁡θ=cos⁡θ/sin⁡θ\operatorname{cot}\theta = \operatorname{cos}\theta \mathbin{/} \operatorname{sin}\theta

Cotangent is cosine divided by sine.

0.5000 / 0.8660 = 0.5774
See proof

Pythagorean

sin⁡2θ+cos⁡2θ=1\operatorname{sin}^2\theta + \operatorname{cos}^2\theta = 1

Pythagoras' theorem in the orange–purple triangle, with hypotenuse 1.

0.7500 + 0.2500 = 1.0000
See proof
1+tan⁡2θ=sec⁡2θ1 + \operatorname{tan}^2\theta = \operatorname{sec}^2\theta

Pythagoras in the big triangle on the line x = 1 (yellow and red).

1 + 3.0000 = 4.0000 = 2.0000²
See proof
1+cot⁡2θ=csc⁡2θ1 + \operatorname{cot}^2\theta = \operatorname{csc}^2\theta

Pythagoras in the triangle on the line y = 1 (blue and green).

1 + 0.3333 = 1.3333 = 1.1547²
See proof
4 · Signs

Sign by quadrant

Sine is positive above the x-axis; cosine, to the right of the y-axis. The others inherit their sign from these two.

II · 90°–180° Positive: sin, csc
I · 0°–90° Positive: all
III · 180°–270° Positive: tan, cot
IV · 270°–360° Positive: cos, sec
5 · Worth memorising

Key angles

Tap a row to move the point to that angle. For other quadrants, use the reference angle and add the quadrant's sign.

DegreesRadianssincostancotseccsc
0010∄1∄
π/61/2√3/2√3/3√32√3/32
π/4√2/2√2/211√2√2
π/3√3/21/2√3√3/322√3/3
π/210∄0∄1
π0−10∄−1∄
3π/2−10∄0∄−1
6 · What it's for

Where they're used

Solving trianglesFinding unknown sides and angles: heights, distances, slopes.
Resolving vectorsIn physics, a force F at angle θ has components F cos θ and F sin θ.
Circular motion and wavesA rotating point traces (cos t, sin t); its height over time is a sine wave (sound, alternating current).
Graphics and rotationsRotating a point in a game or a chart uses a matrix with the cos and sin of the angle.
Proof

Why sin θ = y

sin⁡θ=y\operatorname{sin}\theta = y
  1. P=(x,y)P = (x, y)

    P is where the angle's side crosses the circle. Dropping a vertical line from P to the x-axis forms a right triangle (one of its angles is 90°).

  2. hypotenuse=1\text{hypotenuse} = \hl{1}

    The hypotenuse is the longest side, the one opposite the right angle. Here it's the circle's radius, which is 1.

  3. sin⁡θ=oppositehypotenuse\operatorname{sin}\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}

    Definition of sine. The opposite side is the one facing θ, not touching it: the point's height.

  4. sin⁡θ=y1\operatorname{sin}\theta = \dfrac{\hl{y}}{\hl{1}}

    Substitute: the opposite side is y and the hypotenuse is 1.

  5. sin⁡θ=y\operatorname{sin}\theta = y

    Dividing by 1 changes nothing.

What it's for

You can read the sine straight off the drawing: it's the height (y-coordinate) of P.

Proof

Why cos θ = x

cos⁡θ=x\operatorname{cos}\theta = x
  1. P=(x,y)P = (x, y)

    As with sine, P and its shadow on the x-axis form a right triangle.

  2. hypotenuse=1\text{hypotenuse} = \hl{1}

    The hypotenuse is the circle's radius.

  3. cos⁡θ=adjacenthypotenuse\operatorname{cos}\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}

    Definition of cosine. The adjacent side is the one touching θ (other than the hypotenuse): the triangle's base.

  4. cos⁡θ=x1=x\operatorname{cos}\theta = \dfrac{\hl{x}}{\hl{1}} = x

    The base is x and the hypotenuse is 1.

What it's for

Cosine is P's x-coordinate: how far the point has moved horizontally.

Proof

Why the yellow segment is tan θ

tan⁡θ=sin⁡θcos⁡θ=t\operatorname{tan}\theta = \dfrac{\operatorname{sin}\theta}{\operatorname{cos}\theta} = t
  1. tan⁡θ=oppositeadjacent=yx\operatorname{tan}\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{y}{x}

    Definition of tangent in a right triangle. In the small triangle the opposite side is y and the adjacent side is x.

  2. △(x,y)  ∼  △(1,t)\triangle(x, y) \;\sim\; \triangle(\hl{1}, \hl{t})

    Extend the angle's side to the line x = 1: a big triangle appears with base 1 and height t (the yellow segment). The symbol ∼ means "similar".

  3. t1=yx\dfrac{t}{1} = \dfrac{y}{x}

    Both triangles have the same angles (they share θ and both have 90°). They are similar: same shape, different size, so their sides are proportional.

  4. t=yx=tan⁡θt = \dfrac{y}{x} = \operatorname{tan}\theta

    Solve for t. The yellow segment is exactly the tangent. It's named that because the line x = 1 touches the circle at a single point: it's tangent.

What it's for

Tangent is the slope of the angle's line: how much it rises per unit it moves forward.

Proof

Why the blue segment is cot θ

cot⁡θ=cos⁡θsin⁡θ=c\operatorname{cot}\theta = \dfrac{\operatorname{cos}\theta}{\operatorname{sin}\theta} = c
  1. cot⁡θ=cos⁡θsin⁡θ=1tan⁡θ\operatorname{cot}\theta = \dfrac{\operatorname{cos}\theta}{\operatorname{sin}\theta} = \dfrac{1}{\operatorname{tan}\theta}

    Definition: cotangent is the reciprocal of tangent.

  2. △ (0,0), (0,1), (c,1)\triangle\ (0, 0),\ (0, 1),\ (c, 1)

    Extend the angle's side to the horizontal line y = 1. This forms a triangle with height 1 and base c (the blue segment).

  3. ∠(c,1)=θ\angle(c, 1) = \theta

    The line y = 1 is parallel to the x-axis, and the diagonal crosses both at the same angle (alternate angles).

  4. tan⁡θ=1c\operatorname{tan}\theta = \dfrac{\hl{1}}{\hl{c}}

    From that corner, the opposite side is 1 and the adjacent side is c: tangent = opposite / adjacent.

  5. c=1tan⁡θ=cot⁡θc = \dfrac{1}{\operatorname{tan}\theta} = \operatorname{cot}\theta

    Solve for c.

What it's for

It's the reciprocal of the slope; it shows up when studying nearly vertical lines.

Proof

Why the red segment is sec θ

sec⁡θ=1cos⁡θ=h\operatorname{sec}\theta = \dfrac{1}{\operatorname{cos}\theta} = h
  1. △: 1, t, h\triangle:\ 1,\ t,\ \hl{h}

    It's the same big triangle as for tangent: base 1, height t. Its hypotenuse h is the red segment, from the origin to the line x = 1.

  2. cos⁡θ=adjacenthypotenuse=1h\operatorname{cos}\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{h}

    Definition of cosine applied to the big triangle: the adjacent side is 1 and the hypotenuse is h.

  3. h=1cos⁡θ=sec⁡θh = \dfrac{1}{\operatorname{cos}\theta} = \operatorname{sec}\theta

    Solve for h. That's why |sec θ| is always 1 or more: the hypotenuse is never shorter than the base.

What it's for

It appears in physics (components of inclined forces) and in integrals.

Proof

Why the green segment is csc θ

csc⁡θ=1sin⁡θ=h\operatorname{csc}\theta = \dfrac{1}{\operatorname{sin}\theta} = h
  1. △: c, 1, h\triangle:\ c,\ 1,\ \hl{h}

    It's the cotangent triangle: base c, height 1. Its hypotenuse h is the green segment, from the origin to the line y = 1.

  2. sin⁡θ=oppositehypotenuse=1h\operatorname{sin}\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{h}

    The angle θ sits at the corner (c, 1). From there, the opposite side is 1.

  3. h=1sin⁡θ=csc⁡θh = \dfrac{1}{\operatorname{sin}\theta} = \operatorname{csc}\theta

    Solve for h.

What it's for

Completes the six ratios; useful for simplifying expressions with 1/sin.

Proof

Degrees and radians

θrad=θ°⋅π180\theta_{\text{rad}} = \theta\degree \cdot \dfrac{\pi}{180}
  1. L=2πrL = 2\pi r

    Length L of a circle of radius r.

  2. r=1  →  L=2πr = 1 \;\to\; L = 2\pi

    On the unit circle a full turn has length 2π.

  3. 1 rad: ⌢=r1\ \text{rad}:\ \frown = r

    A radian is the angle spanning an arc (⌒) equal to the radius. That's why a full turn has 2π radians.

  4. 360°=2π rad  →  180°=π rad360\degree = 2\pi\ \text{rad} \;\to\; 180\degree = \pi\ \text{rad}

    The same turn measured two ways. Divide by 2.

  5. 60°⋅π180=π3 rad60\degree \cdot \dfrac{\pi}{180} = \dfrac{\pi}{3}\ \text{rad}

    Rule of three: multiply by π/180 to get radians; by 180/π to go back to degrees.

What it's for

Physics and calculus formulas (e.g. the derivatives of sin and cos) only work in radians.

Proof

Signs by quadrant

sin⁡θ=y⋅cos⁡θ=x\operatorname{sin}\theta = y \quad\cdot\quad \operatorname{cos}\theta = x
  1. sin⁡θ=y,cos⁡θ=x\operatorname{sin}\theta = y,\quad \operatorname{cos}\theta = x

    Sine has the sign of y and cosine the sign of x.

  2. I: x>0, y>0\text{I}:\ x > 0,\ y > 0

    Top right: all positive.

  3. II: x<0, y>0\text{II}:\ x < 0,\ y > 0

    Top left: only sine (and its reciprocal, cosecant) is positive.

  4. III: x<0, y<0\text{III}:\ x < 0,\ y < 0

    Bottom left: sin and cos are negative, but tan = y/x is negative/negative = positive.

  5. IV: x>0, y<0\text{IV}:\ x > 0,\ y < 0

    Bottom right: only cosine (and secant) is positive.

  6. sec⁡∼cos⁡,csc⁡∼sin⁡,cot⁡∼tan⁡\operatorname{sec} \sim \operatorname{cos},\quad \operatorname{csc} \sim \operatorname{sin},\quad \operatorname{cot} \sim \operatorname{tan}

    A ratio and its reciprocal always share their sign (∼), because 1 divided by a number keeps its sign.

What it's for

Work out any angle with its reference angle: table value + quadrant sign.

Proof

Where the key angles come from

sin⁡30°=12⋅sin⁡45°=22⋅sin⁡60°=32\operatorname{sin} 30\degree = \dfrac{1}{2} \quad\cdot\quad \operatorname{sin} 45\degree = \dfrac{\sqrt{2}}{2} \quad\cdot\quad \operatorname{sin} 60\degree = \dfrac{\sqrt{3}}{2}
  1. 45°: a2+a2=1245\degree:\ a^2 + a^2 = 1^2

    If one acute angle is 45°, so is the other (they add up to 90°). Both legs are equal; call them a. Apply Pythagoras with hypotenuse 1.

  2. 2a2=1  →  a=12=222a^2 = 1 \;\to\; a = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}

    Solve for a. Rationalising means multiplying top and bottom by √2 to remove the root from the denominator.

  3. sin⁡45°=cos⁡45°=22\operatorname{sin} 45\degree = \operatorname{cos} 45\degree = \dfrac{\sqrt{2}}{2}

    Both legs have the same length.

  4. 60°+60°+60°=180°  →  30°, 60°, 90°60\degree + 60\degree + 60\degree = 180\degree \;\to\; 30\degree,\ 60\degree,\ 90\degree

    An equilateral triangle (three equal sides) has three 60° angles. Cut in half, it gives two 30°-60°-90° triangles with hypotenuse 1.

  5. b=12,(12)2+h2=1  →  h=32b = \dfrac{1}{2},\quad \left(\dfrac{1}{2}\right)^2 + h^2 = 1 \;\to\; h = \dfrac{\sqrt{3}}{2}

    The base b is split into two halves. The height h comes from Pythagoras: h² = 1 − 1/4 = 3/4.

  6. sin⁡30°=12,cos⁡30°=32,sin⁡60°=32,cos⁡60°=12\operatorname{sin} 30\degree = \dfrac{1}{2},\quad \operatorname{cos} 30\degree = \dfrac{\sqrt{3}}{2},\quad \operatorname{sin} 60\degree = \dfrac{\sqrt{3}}{2},\quad \operatorname{cos} 60\degree = \dfrac{1}{2}

    Opposite 30° is the short base; opposite 60° is the height.

What it's for

With two triangles you can rebuild the whole table without memorising it.

Proof

sin θ · csc θ = 1

sin⁡θ⋅csc⁡θ=1\operatorname{sin}\theta \cdot \operatorname{csc}\theta = 1
  1. csc⁡θ=1sin⁡θ\operatorname{csc}\theta = \dfrac{1}{\operatorname{sin}\theta}

    Definition of cosecant: the reciprocal of sine.

  2. sin⁡θ⋅csc⁡θ=sin⁡θ⋅1sin⁡θ\operatorname{sin}\theta \cdot \operatorname{csc}\theta = \operatorname{sin}\theta \cdot \hl{\dfrac{1}{\operatorname{sin}\theta}}

    Substitute the cosecant.

  3. sin⁡θ⋅csc⁡θ=sin⁡θsin⁡θ=1\operatorname{sin}\theta \cdot \operatorname{csc}\theta = \dfrac{\operatorname{sin}\theta}{\operatorname{sin}\theta} = 1

    A number divided by itself is 1. Only valid if sin θ ≠ 0 (you can't divide by 0).

What it's for

That's why they're called reciprocal ratios: multiplied together they give 1. It lets you switch from one to the other.

Proof

cos θ · sec θ = 1

cos⁡θ⋅sec⁡θ=1\operatorname{cos}\theta \cdot \operatorname{sec}\theta = 1
  1. sec⁡θ=1cos⁡θ\operatorname{sec}\theta = \dfrac{1}{\operatorname{cos}\theta}

    Definition of secant: the reciprocal of cosine.

  2. cos⁡θ⋅sec⁡θ=cos⁡θ⋅1cos⁡θ\operatorname{cos}\theta \cdot \operatorname{sec}\theta = \operatorname{cos}\theta \cdot \hl{\dfrac{1}{\operatorname{cos}\theta}}

    Substitute the secant.

  3. cos⁡θ⋅sec⁡θ=cos⁡θcos⁡θ=1\operatorname{cos}\theta \cdot \operatorname{sec}\theta = \dfrac{\operatorname{cos}\theta}{\operatorname{cos}\theta} = 1

    It simplifies, as long as cos θ ≠ 0.

What it's for

If you know the cosine, the secant is its reciprocal (and vice versa).

Proof

tan θ · cot θ = 1

tan⁡θ⋅cot⁡θ=1\operatorname{tan}\theta \cdot \operatorname{cot}\theta = 1
  1. tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\operatorname{tan}\theta = \dfrac{\operatorname{sin}\theta}{\operatorname{cos}\theta},\quad \operatorname{cot}\theta = \dfrac{\operatorname{cos}\theta}{\operatorname{sin}\theta}

    Write both in terms of sine and cosine.

  2. tan⁡θ⋅cot⁡θ=sin⁡θcos⁡θ⋅cos⁡θsin⁡θ\operatorname{tan}\theta \cdot \operatorname{cot}\theta = \dfrac{\operatorname{sin}\theta}{\operatorname{cos}\theta} \cdot \dfrac{\operatorname{cos}\theta}{\operatorname{sin}\theta}

    Multiply.

  3. tan⁡θ⋅cot⁡θ=sin⁡θ⋅cos⁡θcos⁡θ⋅sin⁡θ=1\operatorname{tan}\theta \cdot \operatorname{cot}\theta = \dfrac{\hl{\operatorname{sin}\theta} \cdot \operatorname{cos}\theta}{\operatorname{cos}\theta \cdot \hl{\operatorname{sin}\theta}} = 1

    To multiply fractions, multiply top by top and bottom by bottom. Top and bottom are the same: 1.

What it's for

Switch from tangent to cotangent with cot θ = 1/tan θ.

Proof

tan θ = sin θ / cos θ

tan⁡θ=sin⁡θcos⁡θ\operatorname{tan}\theta = \dfrac{\operatorname{sin}\theta}{\operatorname{cos}\theta}
  1. tan⁡θ=oppositeadjacent\operatorname{tan}\theta = \dfrac{\text{opposite}}{\text{adjacent}}

    Definition of tangent in the right triangle.

  2. tan⁡θ=yx\operatorname{tan}\theta = \dfrac{y}{x}

    On the unit circle the opposite side is y and the adjacent side is x.

  3. tan⁡θ=sin⁡θcos⁡θ\operatorname{tan}\theta = \dfrac{\hl{\operatorname{sin}\theta}}{\hl{\operatorname{cos}\theta}}

    Because y = sin θ and x = cos θ.

What it's for

It's how you find the tangent when you only know sine and cosine.

Proof

cot θ = cos θ / sin θ

cot⁡θ=cos⁡θsin⁡θ\operatorname{cot}\theta = \dfrac{\operatorname{cos}\theta}{\operatorname{sin}\theta}
  1. cot⁡θ=adjacentopposite\operatorname{cot}\theta = \dfrac{\text{adjacent}}{\text{opposite}}

    Definition of cotangent: the tangent flipped over.

  2. cot⁡θ=xy\operatorname{cot}\theta = \dfrac{x}{y}

    On the unit circle.

  3. cot⁡θ=cos⁡θsin⁡θ\operatorname{cot}\theta = \dfrac{\hl{\operatorname{cos}\theta}}{\hl{\operatorname{sin}\theta}}

    Replace x and y with cosine and sine.

What it's for

Writing everything in sine and cosine is the first step to simplifying almost any expression.

Proof

sin²θ + cos²θ = 1

sin⁡2θ+cos⁡2θ=1\operatorname{sin}^2\theta + \operatorname{cos}^2\theta = 1
  1. a2+b2=c2a^2 + b^2 = c^2

    Pythagoras' theorem: in a right triangle, the square of the hypotenuse (c) is the sum of the squares of the legs (a and b).

  2. a=x,b=y,c=1a = \hl{x},\quad b = \hl{y},\quad c = \hl{1}

    On the unit circle the legs are x and y, and the hypotenuse is the radius.

  3. x2+y2=(1)2x^2 + y^2 = (1)^2

    Substitute.

  4. (cos⁡θ)2+(sin⁡θ)2=1(\operatorname{cos}\theta)^2 + (\operatorname{sin}\theta)^2 = 1

    Replace x with cos θ and y with sin θ.

  5. cos⁡2θ+sin⁡2θ=1\operatorname{cos}^2\theta + \operatorname{sin}^2\theta = 1

    We write cos²θ instead of (cos θ)²; it's just notation.

  6. sin⁡2θ+cos⁡2θ=1\operatorname{sin}^2\theta + \operatorname{cos}^2\theta = 1

    The order of addition doesn't matter.

What it's for

If you know one ratio you get the other: cos θ = ±√(1 − sin²θ). It's the basis of almost every other identity.

Proof

1 + tan²θ = sec²θ

1+tan⁡2θ=sec⁡2θ1 + \operatorname{tan}^2\theta = \operatorname{sec}^2\theta
  1. sin⁡2θ+cos⁡2θ=1\operatorname{sin}^2\theta + \operatorname{cos}^2\theta = 1

    Start from the Pythagorean identity.

  2. sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ\dfrac{\operatorname{sin}^2\theta}{\hl{\operatorname{cos}^2\theta}} + \dfrac{\operatorname{cos}^2\theta}{\hl{\operatorname{cos}^2\theta}} = \dfrac{1}{\hl{\operatorname{cos}^2\theta}}

    Divide every term by cos²θ (allowed if cos θ ≠ 0). Doing the same to both sides keeps the equality.

  3. tan⁡2θ+1=sec⁡2θ\operatorname{tan}^2\theta + 1 = \operatorname{sec}^2\theta

    sin/cos is tan, cos/cos is 1 and 1/cos is sec.

  4. 1+tan⁡2θ=sec⁡2θ1 + \operatorname{tan}^2\theta = \operatorname{sec}^2\theta

    Reorder. In the figure it's Pythagoras in the big triangle: legs 1 and tan θ, hypotenuse sec θ.

What it's for

Go from tangent to secant without finding the angle; it shows up a lot in integrals.

Proof

1 + cot²θ = csc²θ

1+cot⁡2θ=csc⁡2θ1 + \operatorname{cot}^2\theta = \operatorname{csc}^2\theta
  1. sin⁡2θ+cos⁡2θ=1\operatorname{sin}^2\theta + \operatorname{cos}^2\theta = 1

    Start from the Pythagorean identity.

  2. sin⁡2θsin⁡2θ+cos⁡2θsin⁡2θ=1sin⁡2θ\dfrac{\operatorname{sin}^2\theta}{\hl{\operatorname{sin}^2\theta}} + \dfrac{\operatorname{cos}^2\theta}{\hl{\operatorname{sin}^2\theta}} = \dfrac{1}{\hl{\operatorname{sin}^2\theta}}

    Now divide by sin²θ (if sin θ ≠ 0).

  3. 1+cot⁡2θ=csc⁡2θ1 + \operatorname{cot}^2\theta = \operatorname{csc}^2\theta

    sin/sin is 1, cos/sin is cot and 1/sin is csc. It's Pythagoras in the triangle on the line y = 1.

What it's for

Links cotangent and cosecant just as the previous one links tangent and secant.

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