Why is $\sin(x) = \sin(180^{\circ}-x)$ I cannot seem to understand why this is true. Same for $\cos(x) = -\cos(180^{\circ}-x)$ and $\tan(x) = -\tan(180^{\circ}-x)$. Without the use of the compound angle formulas. 
Thanks
 A: Hint  Rebember that 
$
\tan( 180^o \pm x  ) = \frac{\sin( 180^o \pm x)}{\cos(180^o\pm x)}.
$
Now draw the  trigonometric cycle and observes the measures of  sine and cosine.

A: There are geometric reasons for the relations $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)= -\cos x$ (I prefer not using degrees, change $\pi$ into degrees, if you want).
The historic definition of sine and cosine are by means of rectangle triangles. If $ABC$ is a triangle with a right angle in $B$ and $\alpha$ is the angle with vertex in $A$, then
$$
\sin\alpha=\frac{BC}{AC},\quad
\cos\alpha=\frac{AB}{AC},
$$
so that the relation $\sin^2\alpha+\cos^2\alpha=1$ follows from Pythagoras' theorem.  It follows also, by the very definition, that
$$
\sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha,\quad
\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha.
$$
This defines the sine and the cosine for all acute angles. What can we do for obtuse angles?
Consider the triangle in the following figure

If we call $a=BC$, $b=AC$, $c=AB$ (I omit the bar to denote the length for simplicity) and $d=AH$ (where $CH$ is the perpendicular to $AB$), we can easily see by a double application of Pythagoras' theorem that
$$
a^2=b^2+c^2-2cd
$$
or, introducing the cosine,
$$
a^2=b^2+c^2-2bc\cos\alpha.
$$
Suppose now we have an obtusangle triangle

With the same notation as before, we have
$$
a^2=b^2+c^2+2cd
$$
or, introducing the cosine
$$
a^2=b^2+c^2+2bc\cos(\pi-\alpha)
$$
because now $d=b\cos(\pi-\alpha)$. By defining
$$
\cos\alpha=-\cos(\pi-\alpha)
$$
for an obtuse angle $\alpha$, the relation
$$
a^2=b^2+c^2-2bc\cos\alpha
$$
(al-Kashi theorem) holds for all triangles. It only remains to supplement the definition of $\cos(\pi/2)$, but by Pythagoras' theorem we need to set
$$
\cos\frac{\pi}{2}=0
$$
For the sine, the reasoning can start from the sine's theorem:

The figure, with some simple reasoning, gives
$$
a\sin\alpha=2R
$$
where $R=OB=OC=OA$. If you take the diametral opposite point $A'$ and  of $A$, you immediately get that the sine of the supplementary angle $\alpha'$ of $\alpha$ must obey, if its sine were defined,
$$
a\sin\alpha'=2R
$$
that means
$$
\sin(\pi-\alpha)=\sin\alpha.
$$
Again, this forces us to set $\sin(\pi/2)=1$, which agrees with $\cos(\pi/2)=0$. For the zero angle, we can use the complementary angle rules to conclude we have to define $\sin0=0$ and $\cos0=1$.
A: Draw a circle in the Cartesian plane with the centre at $0$ and see that $\sin x$ is the $y$ axis and figure out that $\sin x$ is positive in the the upper part of the unit disk.
A: Because, if you take $\cos(180°-x)$ you are effectively getting the opposite of $\cos(x)$. Due to $180°$ being halfway over the circle diameter. So if you do take the opposite of that $-\cos(180°-x)$ you get $x$ back, because you switch the sign.
Easy to see with this image:
