# Proving $\sqrt{\left(\cos t-1\right)^2+\sin^2t} = 2\,\left|\sin\frac{t}{2}\right|$

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".

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

$$\cos\left(\frac t2+\frac t2\right)=\cos\left(\frac t2\right)\cos\left(\frac t2\right)-\sin\left(\frac t2\right)\sin\left(\frac t2\right)=\cos^2\left(\frac t2\right)-\sin^2\left(\frac t2\right).$$
Therefore, $$(\cos(t)-1)^2+(\sin(t))^2=\cos^2(t)-2\cos(t)+1+\sin^2(t)=2-2\cos(t)=$$
$$2-2\cos^2\left(\frac t2\right)+2\sin^2\left(\frac t2\right)=2\sin^2\left(\frac t2\right)+2\sin^2\left(\frac t2\right)=(2 \sin\left(\frac t2)\right)^2,$$