Why is the sine/cosine of an angle equal to its supplement? I just got my hands on a trig text and I've been studying the law of sines and cosines so I can solve triangles other than right triangles. Something I've found odd while studying proofs of these theorems are the statements that the sine/cosine of an angle is equal to its supplement. This does not seem intuitive to me and I'm having a hard time understanding how the sine of a 45 degree angle can equal the sine of a 135 degree angle. Can someone please explain this concept to me?
Thanks
 A: It  is not true for cosine.
$\cos (\theta) = -\cos (180-\theta)$
The most simpleminded way to learn these functions is to start from the unit circle.  Draw a circle of radius 1 centered at the origin.  Draw in a radius.  $\theta$ is the angle between the positive x-axis and your radius, measured counterclockwise from the positive x-axis.
The y-coordinate where this radius intersects the circle is $\sin\theta$,  the x-coordinate is $\cos\theta$.  
Starting from this framework, it should be a little bit more clear that $\sin(180-\theta)=\sin(\theta)$.
If you are starting from the law of sines.  If you have $\triangle ABC$, then the area of $\triangle ABC = (mAB)(mAC)\sin A$
If you construct a point D such that $\angle DAC$ is supplementary to $\angle BAC$ and $mAD = mAB$, then DB is parallel to AC, and Area $\triangle DAC$ = Area $\triangle BAC$.
area of $\triangle ABC = (mAB)(mAC)\sin A$ = area of $\triangle ADC = (mAD)(mAC)\sin \sup A$
$\sin A = \sin \sup A$
A: The sine has this property; the cosine has a slightly more complicated property:
\begin{align}
\sin(\pi-\theta) & = \sin\theta, \\[6pt]
\cos(\pi-\theta) & = -\cos\theta.
\end{align}
This is best understood by looking at the unit-circle characterizations of sine and cosine:



A: As with many things in trig, there is more than one way to approach this.  I don't know your trig background so I'll be as basic as possible.
The unit circle approach
The unit circle is the circle with center $(0,0)$ and radius $1$.  dreamer's comment to your original question contains a good picture.  In terms of the unit circle, $\cos\theta$ and $\sin\theta$ are defined to be the $x$- and $y$-coordinate, respectively, of the point of intersection the (terminal side of the) angle $\theta$ makes with the unit circle.  From the picture, you can see that an angle of $45^\circ$ hits the unit circle at the point $\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$.  So the cosine of $45^\circ$ is the $x$-coordinate of this point, and the sine of $45^\circ$ is the $y$-coordinate of this point.  Similarly, we see from the picture that an angle of $135^\circ$ hits the unit circle at the point $\left(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$.
The trig identity approach
Here are two well-known trig identities:
\begin{align}
  \cos(a-b) &= \cos a\cos b + \sin a \sin b\\
  \sin(a-b) &= \sin a\cos b - \cos a \sin b
\end{align}
Let $a = 180^\circ$ in those identities (and let $b$ just represent any arbitrary angle) to get the desired results:
\begin{align}
  \cos(180^\circ - b) &= \cos 180^\circ \cos b + \sin 180^\circ \sin b\\
    &= (-1) \cdot \cos b + 0 \cdot \sin b\\
    &= -\cos b\\
  \sin(180^\circ - b) &= \sin 180^\circ \cos b - \cos 180^\circ \sin b\\
    &= 0 \cdot \cos b - (-1) \cdot \sin b\\
    &= \sin b
\end{align}
Here we've used the facts that $\cos 180^\circ = -1$ and $\sin 180^\circ = 0$.  These can be easily obtained from the unit circle.
If any of this is over your head, don't worry about it.  As you continue your studies in trig it will start to make sense and you'll start to see how a lot of these properties relate to each other.
Shameless self-plug:  I do have a series of trig tutoring videos on YouTube if you want to learn more.
