Answer:
To prove, \[\sin \frac{\theta }{2}=\frac{1}{2}|\hat{a}-\hat{b}|\] We know that, \[|\hat{a}-\hat{b}{{|}^{2}}=(\hat{a}-\hat{b})\cdot (\hat{a}-\hat{b})\] \[=\hat{a}\cdot \hat{a}\,-\hat{a}\cdot \hat{b}\,-\hat{b}\cdot \hat{a}\,+\hat{b}\cdot \hat{b}\] \[=\,\,|\hat{a}{{|}^{2}}-2\,\hat{a}\cdot \hat{b}+|\hat{b}{{|}^{2}}\] [\[\because \] dot product is commutative] \[=1-2\hat{a}\cdot \hat{b}+1\] \[[\because \,\,\,|\hat{a}|\,\,=\,\,|\hat{b}|\,\,=1]\] \[\Rightarrow \] \[|\hat{a}-\hat{b}{{|}^{2}}=2-2\cos \theta \] \[\begin{align} & [\text{if}\,\,\hat{a}\cdot \hat{b}=\,\,|\hat{a}||\hat{b}|\,\,\cos \theta \,\,\text{and}\,\,|\hat{a}|\,\,=\,\,|\hat{b}|\,\,=1, \\ & then\,\,\hat{a}\cdot \hat{b}=\cos \theta ] \\ \end{align}\] \[=2\,(1-\cos \theta )=2\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)=4{{\sin }^{2}}\frac{\theta }{2}\] \[\left[ \because \,\,\,1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} \right]\] \[\Rightarrow \] \[{{\sin }^{2}}\frac{\theta }{2}=\frac{1}{4}|\hat{a}-\hat{b}{{|}^{2}}\] \[\therefore \] \[\sin \frac{\theta }{2}=\frac{1}{2}|\hat{a}-\hat{b}|\] [Taking positive square root as modulus cannot be negative] Hence proved.
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