Help me derive this basic trigonometry identity

Question

$$-\sin^2(x)+1 = \cos^2(x)$$ I saw this identity and I want to know how you derive it.

Answer 1

By Pythagoras theorem,

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

$(\frac{a}{c})^2+(\frac{b}{c})^2=1$

$\cos^2x + \sin^2x=1$

or

$$1-\sin^2x=\cos^2x$$

Answer 2

We wish to prove that $\sin^2(x) + \cos^2(x) = 1$. Consider a right-angled triangle with one of the angles as $x$. Now $sin^2(x) = \frac{o^2}{h^2}, cos^2(x) = \frac{a^2}{h^2}$ where $o, a, h$ are the opposite side, adjacent side, and the hypothenuse. But $o^2 + a^2 = h^2$ by Pythagoras's theorem, so $\sin^2(x) + \cos^2(x) = \frac{o^2+a^2}{h^2} = 1$.

Answer 3

This is called the trigonometric Pythagoras. I give you two ways to obtain the identity.

Method 1: We use the Euler-identity $\mathrm{e}^{\mathrm{i} \varphi} = \cos(\varphi) + \mathrm{i} \sin (\varphi)$ for every $\varphi \in \mathbb{R}$. With this we have

$$1 = \mathrm{e}^0 = \mathrm{e}^{\mathrm{i}\varphi - \mathrm{i}\varphi} = \mathrm{e}^{\mathrm{i}\varphi}\mathrm{e}^{-\mathrm{i}\varphi} = \big(\cos(\varphi) + \mathrm{i} \sin(\varphi)\big)\big(\cos(-\varphi) + \mathrm{i} \sin(-\varphi)\big) = \cos^2(\varphi) + \sin^2(\varphi)$$

since $\cos(-\varphi) = \cos(\varphi)$ and $\sin(-\varphi) = - \sin(\varphi)$.

Method 2: Define the real - valued function $f(x) := \sin^2(x) + \cos^2(x)$ for every $x \in \mathbb{R}$. Then $f$ is differentiable with derivative

$$f'(x) = 2\sin(x)\cos(x) - 2\sin(x)\cos(x) \equiv 0,$$

hence $f$ must be constant: $f \equiv c$ with an unknown $c \in \mathbb{R}$. For knowing the value of a constant function everywhere, it clearly suffices to know the value at one point. Since $f(0) = \cos^2(0) + \sin^2(0) = 1$, it follows $f \equiv 1$. We obtain

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