Can someone please explain how and why $\sin(\pi - a) = \sin (a)$? The handout from class doesn't really explain it. I tried asking the teach but there's a bit of a language barrier and I'm not getting how he's explaining it.
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$\begingroup$First, use the fact that $\sin$ is an odd function, so $\sin (\pi -a)=-\sin (a-\pi)$. Now, shifting a sine or cosine function by half a period flips the sign again (You can look at the graph to see the reason why), so $\sin (a-\pi)=-\sin (a)$. Combining the two together, the double negatives cancel and we get the desired equality
$\endgroup$ 2 $\begingroup$Looking at it in terms of the unit circle, $\sin(a)$ represents the $y$-value when you travel counterclockwise $a$ radians, starting from the point $(1,0)$. $\sin(\pi-a)$ then would represent you starting halfway around ($\pi$ radians) the unit circle, at the point $(-1,0)$, and then going $a$ radians clockwise, since it is negative. If you look at those two points at any angle $a$, you see that one is essentially just a reflection across the $y$-axis of the other, and so while the $x$-value is not the same (negative), the $y$-value will still be the same.
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