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

Given \(\mathbf{a}\) and \(\mathbf{b}\) are unit vectors
\(\therefore|\mathbf{a}|=1 \text { and }|\mathbf{b}|=1\)
We know that, \(|\mathbf{a}-\mathbf{b}|^{2}=(\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})\)
\(=|\mathbf{a}|^{2}-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^{2}\)
\(=|\mathbf{a}|^{2}-2|\mathbf{a}||\mathbf{b}| \cos \theta+|\mathbf{b}|^{2}\)
\(=1-2 \cdot 1 \cdot 1 \cdot \cos \theta+1\)
\(=2-2 \cos \theta\)
\(\Rightarrow \quad|\mathbf{a}-\mathbf{b}|^{2}=2(1-\cos \theta)\)
\(=2\left(2 \sin ^{2} \theta / 2\right)\)
\(=4 \sin ^{2} \theta / 2\)
\(\begin{array}{ll}\Rightarrow & \sin ^{2} \frac{\theta}{2}=\frac{1}{4}|\mathbf{a}-\mathbf{b}| \\ \Rightarrow & \sin \frac{\theta}{2}=\frac{1}{2}|\mathbf{a}-\mathbf{b}|\end{array}\)