notice the picture below
x = 1 and y = -1
now... recall your SOH CAH TOA \(\bf sin(\theta)=\cfrac{y}{r}
\qquad
% cosine
cos(\theta)=\cfrac{x}{r}
\qquad
% tangent
tan(\theta)=\cfrac{y}{x}\)
now.. what's the value for "r" or the radius? well, using the pythagorean theorem \(\bf c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad
\begin{cases}
a=x\\
b=y\\
c=r
\end{cases}\)
