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:

\[\forall x\in \mathbb{R}, \quad\cos^2 x+ \sin^2 x =1\]

Proof/Demonstration using the derivative

Let $f$ be the function defined as follows:

\[\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x\] \[\begin{aligned} f'(x)&=(\cos^2 x+ \sin^2 x)' \\ &= 2 \cos x (-\sin x) + 2 \sin x \cos x\\ &= - 2 \cos x \sin x + 2 \sin x \cos x\\ & =0 \end{aligned}\]

This means that $f$ is constant on $\mathbb{R}$:

\[\exists C \in \mathbb{R},\forall x\in \mathbb{R}, \quad f(x)=C\]

Let’s take $x=0$:

\[f(x=0)=\cos^2 0+ \sin^2 0=1\]

We conclude:

\[\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x =1\]

Proof/Demonstration using addition formulas

We had previously demonstrated the addition formula

\[\cos(a+b)=\cos a \cos b - \sin a \sin b\]

Let $a=x\in \mathbb{R}$. Let $b=-a=-x$, we have, since cosine is an even function and sine an odd function:

\[\begin{aligned} \cos(x-x)&=\cos x \cos (-x) - \sin x \sin (-x)\\ &= \cos x \cos x - \sin x (-\sin x)\\ &= \cos^2 x + \sin^2 x\\ &=\cos 0\\ &=1 \end{aligned}\]

We conclude then:

\[\forall x\in \mathbb{R}, \quad f(x)=\cos^2 x+ \sin^2 x=1\]

Proof/Demonstration using the trigonometric circle

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

\[\begin{aligned} x^2+y^2=r^2&=1 \\ \cos^2 \theta+ \sin^2 \theta&=1 \end{aligned}\]

Then we have:

\[\forall \theta\in [0, 2\pi ], \quad \cos^2 \theta+ \sin^2 \theta =1\]

We conclude that:

\[\forall x\in \mathbb{R}, \quad\cos^2 x+ \sin^2 x =1\]

The conversion from radians $\theta$ to degrees $x$ is done as follows:

\[x=\theta \times \frac{180}{\pi}\]

Example: $\theta=\pi/4$

\[x= \frac{\pi}{4} \times \frac{180 }{\pi}= \frac{180 }{4}=45^\circ\]