We will demonstrate the following equality cos²x + sin²x=1 in several ways. By using the notion of derivative, addition formula and then geometrically by using the trigonometric circle.

Let’s prove the following equality:

∀x∈R,cos2⁡x+sin2⁡x=1

Proof/Demonstration using the derivative

Let f be the function defined as follows:

∀x∈R,f(x)=cos2⁡x+sin2⁡x f′(x)=(cos2⁡x+sin2⁡x)′=2cos⁡x(−sin⁡x)+2sin⁡xcos⁡x=−2cos⁡xsin⁡x+2sin⁡xcos⁡x=0

This means that f is constant on R:

∃C∈R,∀x∈R,f(x)=C

Let’s take x=0:

f(x=0)=cos2⁡0+sin2⁡0=1

We conclude:

∀x∈R,f(x)=cos2⁡x+sin2⁡x=1

Proof/Demonstration using addition formulas

We had previously demonstrated the addition formula

cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b

Let a=x∈R. Let b=−a=−x, we have, since cosine is an even function and sine an odd function:

cos⁡(x−x)=cos⁡xcos⁡(−x)−sin⁡xsin⁡(−x)=cos⁡xcos⁡x−sin⁡x(−sin⁡x)=cos2⁡x+sin2⁡x=cos⁡0=1

We conclude then:

∀x∈R,f(x)=cos2⁡x+sin2⁡x=1

Proof/Demonstration using the trigonometric circle

Consider the trigonometric circle of radius r=1. In the following triangle, we can apply the Pythagorean theorem: x=cos⁡θ, y=sin⁡θ. The hypotenuse r=1, we then have:

x2+y2=r2=1cos2⁡θ+sin2⁡θ=1

Then we have:

∀θ∈[0,2π],cos2⁡θ+sin2⁡θ=1

We conclude that:

∀x∈R,cos2⁡x+sin2⁡x=1

The conversion from radians θ to degrees x is done as follows:

x=θ×180π

Example: θ=π/4

x=π4×180π=1804=45∘