It's trigo time #1

Proving this equation:

$cos^2(-θ)+sin^2(-θ)=1$

$(cos(-θ))^2+(sin(-θ))^2=1$

$cos(-θ)^2+sin(-θ)^2=1$

Then, we can use the Odd and Even identity:

$cos^2θ+sin^2θ=1$

This is also stated by the Pythagorean identity. Let's prove it further:

$1-sin^2θ+sin^2θ=1$

which gives the result:

$\boxed{1=1}$

Hence, we have proved it!

$\begin{aligned} \cos^2 (-\theta) + \sin^2 (-\theta) & = \left(\cos (-\theta)\right)^2 + \left(\sin (-\theta)\right)^2 \\ & = \left(\cos \theta \right)^2 + \left(- \sin \theta \right)^2 \\ & = \cos^2 \theta + \sin^2 \theta \\ & = \boxed{1} \end{aligned}$

Cosine is an even function, meaning that cos(-x) = cos(x). Sine is odd which means that sin(-x) = -sin(x). Both are squared. Cos(x) squared is just cos^2(x) (This is a notational trick to avoid confusion between cos(x)^2 and cos(x^2)). (-sin(x)) squared is also sin^2(x) because the negatives cancel each other out. This leaves us with cos^2(x) + sin^2(x). The Pythagorean Identity tells us that this is equal to 1.

To go back to first principle, we can prove it this way:

Considering a right triangle with sides $a, b, c$ such that $a^2+b^2=c^2$ , we see that $cos(\theta)=\frac{a}{c}$ while $sin(\theta)=\frac{b}{c}$ where $\theta$ is the angle between $a$ and $c$ .

Then $cos^2(-\theta)+sin^2(-\theta)=(\cos(\theta))^2+(-sin(\theta))^2=\cos(\theta))^2+(sin(\theta))^2=(\frac{a}{c})^2+(\frac{b}{c})^2=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=c^0=\boxed{1}$