How can trigonometric functions be negative? I cannot understand why $\cos(180-\theta)$  say is $-\cos\theta$. This is probably because my teacher first introduced trigonometry in triangles. I do not understand it for obtuse angles because I cannot think of them in a right triangle. 
I realised that I couldn't feel what I had read "in my spleen" when I was looking at the proof for the law of cosines in an obtuse-angled triangle. I have spent quite some time thinking about how the "$-\cos\theta$" entered the derivation. I cannot fully understand, why the negatives which work in the $XY$-plane work in triangles. For instance, since in a triangle, all the sides are positive while taking the ratio of sides we do not get any negative values but how then does $\cos 120^{\circ}=-0.5$. My brain is in a mess right now. I would appreciate it if someone could help me out or suggest something that I can do.
Let me illustrate what I can't get around.

It is given that in the triangle $\angle BAC=120$ degrees,$|AC|=3$ and that D is the foot of the perpendicular from C to BD. Then $\cos\angle BAC=-0.5=\dfrac{AD}{AC} \implies AD=-1.5$ ?
 A: The cosine of an obtuse angle simply does not come from ratios of the lengths of the sides of an obtuse angle. It's defined to be the $x$ coordinate of the intersection of the terminal side of the angle with the unit circle. That's all. There's nothing forcing us to make this definition, except that it's immensely useful and agrees with the ratios of sides definition for acute angles. From this definition you can prove that $\cos(180-\theta)=-\cos(\theta)$, using pictures like the one you displayed.
A: Length can be negative.
How is that possible?
Because the negative and positive signs are always relative.
What do I mean by that?
In your setup, you start with A as origin (0,0). Anything on the left-hand side of this is a negative value. If you consider the triangle DCB and ignore rest. Then point D would become (0,0) in that setup and all values to the right-hand side would be positive and all values to the left of D would be negative. All ratios in that DCB triangle will be positive, try out the calculations.
The negative sign essentially denotes the direction. You select a point and assume that all values on one side would be positive and all values on the other side would be negative. It all sounds weird at first, but just practice more and one day it will all start making sense and your mind will actually be blown.
A: For right angle triangles, we all know that values of $sin (\theta)$, $cos (\theta)$ and $tan (\theta)$ works when $0\le\theta\le90$.
We do not touch any angles beyond 90 in right angle triangle trigonometry as total angles in a triangle is 180 degrees.
Now consider these triangles (not to scale):

Where triangle A and Triangle B have different angles labelled.
And we name angle c=$\theta$ and angle d=$(180-\theta)$
Now notice that the cosine rule is: $C^2=B^2+A^2-2\cdot A\cdot B\cdot cos(C)$ for any non-right angle triangle like this one:

Now back to the topic. Notice that for those two triangles (Triangle A and Triangle B), the side length D is clearly bigger than C.
Using the cosine rule for those two triangles:
For triangle A: $C^2=A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)$ (as $c=\theta$)
For Triangle B: $D^2=A^2+B^2-2\cdot A\cdot B\cdot cos(180-\theta)$ (as $d=180-\theta)$
Now you know that $cos(180-\theta) =-cos(\theta)$, we can have for Triangle B: $D=\sqrt{A^2+B^2-(2\cdot A\cdot B\cdot -cos(\theta))}=\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$
Compared to triangle A for side C, $C=\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$
Now we know that the formula for both triangles A and B that the side D is bigger than side C and the formulas derived are:
$C=\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$ for angle $c=\theta$ 
$D=\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$ for $d=180-\theta$
And we make sense that $\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$ is clearly bigger than $\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$
That is one of the reasons why $cos(180-\theta)=-cos(\theta)$
A: Let $\triangle OAB$ denote a right-angled triangle with its right angle at $O$, and define $\theta:=\angle AOB$. If $OB=1$ then $$OA=\cos\theta,\,AB=\sin\theta\left(\ast\right).$$ In the circle of radius $1$ and centre $O$, hold the diameter containing $OA$ fixed as $B$ moves around the circle. (The point $A$ moves along the diameter, but the diameter itself doesn't change.) Now you can define $\cos\theta,\,\sin\theta$ in terms of $\left(\ast\right)$ without thinking about triangles at all.
If $\theta$ is acute then $B$ has Cartesian coordinates $\left(\cos\theta,\,\sin\theta\right)$, provided we set $O=\left(0,\,0\right)$ and identify the half-line $OA$ extended infinitely through $A$ with the positive $x$-axis. (See this diagram.) Now let's increase $\theta$ by continuing to move $B$ anticlockwise. For obtuse $\theta$ you'd have $x<0,\,y>0$. Replacing $\theta$ with $180^\circ-\theta$ reflects $B$ in the radius that ends at $\left(0,\,1\right)$. This replaces $\left(x,\,y\right)$ with $\left(-x,\,y\right)$, which is why $\cos$ changes sign but $\sin$ does not.
