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(C) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$. We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$

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$\frac{|a-b|}{2}$

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(C) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$. We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$. Since $a \cdot b = |a||b| \cos 	heta$,we have $|a-b|^2 = 1^2 - 2(1)(1) \cos 	heta + 1^2 = 2 - 2 \cos 	heta$. Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$,we get $|a-b|^2 = 2(2 \sin^2 \frac{	heta}{2}) = 4 \sin^2 \frac{	heta}{2}$. Taking the square root on both sides,we get $|a-b| = 2 \sin \frac{	heta}{2}$. Therefore,$\sin \frac{	heta}{2} = \frac{|a-b|}{2}$.

If $a$ and $b$ are unit vectors and $\theta$ is the angle between $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

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