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How do I prove that $$\sqrt{\left(\cos t-1\right)^2+\sin^2t} = 2\,\left|\sin\frac{t}{2}\right|$$ and why is it true?

Background:

I was looking at how distances between two points on a circle relate to the arc distance between the two points. I started with points that I knew ,$(-1, 0)$ and $(1, 0)$, whose radians are $\pi$ and $0$. The length of the arc is $\pi$, and the distance between the two points is $2$.

I wasn't sure where to go from there, so I took out my graphing calculator and switched it over to parametric equations. I entered:

$$\left(t, \sqrt{\left(\cos t-1\right)^2+\sin^2t}\right)$$

This gives the distance between a point on the unit circle at radian $t$ and $(1, 0)$. As expected the distance goes up, then down, and never goes below zero. The maximum distance is $2$ and the minimum distance is $0$.

The hill-like pattern reminded me of trig functions, so I did a little bit of thinking and came up with:

$$\left(t, 2\,\left|\sin\frac{t}{2}\right|\right)$$

As expected, it gave me the same result. I looked at it for a second and I wondered why the two were equal. I did a little bit of research and asking around, but the best I got was "prove it".

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  • $\begingroup$ Start by squaring both sides; do you know the double angle formula? $\endgroup$ – J. W. Tanner Apr 30 '20 at 23:31
  • $\begingroup$ @J.W.Tanner I don't, I am a trig noob. $\endgroup$ – DMVerfurth Apr 30 '20 at 23:32
  • $\begingroup$ Do you know angle addition formula? $\endgroup$ – J. W. Tanner Apr 30 '20 at 23:33
  • $\begingroup$ @J.W.Tanner I don't. I know how to use sin and cos, but I'm not that great at manipulating them. $\endgroup$ – DMVerfurth Apr 30 '20 at 23:35
  • $\begingroup$ You know $\sin^2(t)+\cos^2(t)=1$? $\endgroup$ – J. W. Tanner Apr 30 '20 at 23:44
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According to the angle addition formula,

$\cos\left(\frac t2+\frac t2\right)=\cos\left(\frac t2\right)\cos\left(\frac t2\right)-\sin\left(\frac t2\right)\sin\left(\frac t2\right)=\cos^2\left(\frac t2\right)-\sin^2\left(\frac t2\right).$

Therefore, $(\cos(t)-1)^2+(\sin(t))^2=\cos^2(t)-2\cos(t)+1+\sin^2(t)=2-2\cos(t)=$

$2-2\cos^2\left(\frac t2\right)+2\sin^2\left(\frac t2\right)=2\sin^2\left(\frac t2\right)+2\sin^2\left(\frac t2\right)=(2 \sin\left(\frac t2)\right)^2,$

which is essentially what you found.

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