Pythagorean Identities

One Identity You Already Met

Back on the unit circle, every point has coordinates (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and every point sits exactly distance 1 from the center.

That single fact gives us the most important identity in trigonometry.

No matter where the point travels, the green piece and the yellow piece always fill the bar completely. The sum never moves off 1.

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

This holds for every angle. Always.

Why It’s Always True

Look at the triangle inside the circle. It has:

  • a horizontal leg of length cos⁡θ\cos\theta
  • a vertical leg of length sin⁡θ\sin\theta
  • a hypotenuse (the radius) of length 11

That is just a right triangle, so the Pythagorean theorem applies:

(cos⁡θ)2+(sin⁡θ)2=12(\cos\theta)^2 + (\sin\theta)^2 = 1^2

The Pythagorean identity is nothing more than a2+b2=c2a^2 + b^2 = c^2 living on the unit circle. That is why it carries the name.

A Note on the Notation

cos⁡2θ\cos^2\theta means (cos⁡θ)2(\cos\theta)^2: take the cosine first, then square it.

cos⁡2θ=(cos⁡θ)2\cos^2\theta = (\cos\theta)^2, not cos⁡(θ2)\cos(\theta^2).

The little 2 sits up high only to keep the writing tidy. It always squares the whole function value.

Two More, For Free

Once you have the first identity, the other two come out by simple division.

Divide every term by cos⁡2θ\cos^2\theta:

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

Each piece simplifies. Remember tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} and sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}:

1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

Divide every term by sin⁡2θ\sin^2\theta instead:

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

Using cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta} and csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}:

1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

The Three Together

IdentityComes from
cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1the unit circle
1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\thetadivide the first by cos⁡2θ\cos^2\theta
1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\thetadivide the first by sin⁡2θ\sin^2\theta

You only need to remember the first one. The other two are one division away.

What They’re For

These identities let you swap one function for another. If you know one value, you can find the rest.

Example: Suppose sin⁡θ=35\sin\theta = \frac{3}{5} and θ\theta is acute (between 0° and 90°). Find cos⁡θ\cos\theta.

Start from the identity and solve for cos⁡2θ\cos^2\theta:

cos⁡2θ=1−sin⁡2θ=1−925=1625\begin{aligned} \cos^2\theta &= 1 - \sin^2\theta \\[0.5em] &= 1 - \frac{9}{25} \\[0.5em] &= \frac{16}{25} \end{aligned}

Take the square root: cos⁡θ=45\cos\theta = \frac{4}{5}.

We choose the positive root because θ\theta is acute, where cosine is positive. The identity gives the size; the quadrant tells you the sign.