The Pythagorean trigonometric identity – sin^2(x) + cos^2(x) = 1

A very useful and important theorem is the pythagorean trigonometric identity. To understand and prove this theorem we can use the pythagorean theorem.

The trigonometric Identity

$sin^2(x) + cos^2(x) = 1$

You can also write it as

$ (sin(x))^2+(cos(x))^2=1 $

Sine, Cosine and Tangent

A good start to understanding the theorem and to be able to prove it is to know how sine (sin), cosine (cos) and tangent (tan) is defined. They are all relationships between the angles and sides on a right triangle.

Sides of a right triangle

$ sin v = \frac{\text{opposite}}{\text{hypotenuse}} $

$ cos v = \frac{\text{adjacent}}{\text{hypotenuse}} $

$ tan v = \frac{\text{opposite}}{\text{adjacent leg}} $

It´s also important to know that we can change which angle that is designated to be $v$. If we let the upper angle be $v$ the opposite and adjecent sides will switch with each other.

Proof of The Pythagorean trigonometric identity

To prove that $sin^2(x) + cos^2(x) = 1$ we can start by drawing a right triangle.

proof step1

From the pythagorean theorem we know that

$ a^2+b^2=c^2 $

Now lets express $a$ and $b$ by using the sine and cosine.

$ sinv = \frac{b}{c} $
$ b = c·sinv $

$ cosv = \frac{a}{c} $
$ a = c·cosv $

Lets use 1) and 2) in the pythagoreans theorem instead of $a$ and $b$.

$ (c·cosv)^2+(c·sinv)^2=c^2 $

Expand the two parenthesis

$ c^2·cos^2v+c^2·sin^2v=c^2 $

Divide each term on both sides by $c^2$

$\frac{c^2·cos^2v}{c^2}+\frac{c^2·sin^2v}{c^2}=\frac{c^2}{c^2} $

$ cos^2v+sin^2v=1 $

And we are done with the proof!