Question

If $\hat{a}\mathrm{and}\hat{b}$are unit vectors inclined at an angle θ, prove that
(i) $\cos \frac{\mathrm{\theta }}{2}=\frac{1}{2}\left|\hat{a}+\hat{b}\right|$

(ii) $\tan \frac{\mathrm{\theta }}{2}=\frac{\left|\hat{a}-\hat{b}\right|}{\left|\hat{a}+\hat{b}\right|}$

Answer 1

$$\begin{gathered} \text{Given that}\overbrace{a}\text{and}\overbrace{b}\text{are unit vectors.} \\ \text{So,}\left|\hat{a}\right|\text{=1,}\left|\hat{b}\right|\text{=1} \\ \text{We have} \\ {\left|\hat{a}+\hat{b}\right|}^2={\left|\hat{a}\right|}^2+{\left|\hat{b}\right|}^2+2\hat{a}.\hat{b} \\ =1+1+2\left|\hat{a}\right|\left|\hat{b}\right|\cos \theta \\ =2+2\cos \theta \\ \Rightarrow \cos \theta =\frac{{\left|\hat{a}+\hat{b}\right|}^2-2}{2}...\left(1\right) \\ {\left|\hat{a}-b\right|}^2={\left|\hat{a}\right|}^2+{\left|\hat{b}\right|}^2-2\hat{a}.\hat{b} \\ =1+1-2\left|\hat{a}\right|\left|\hat{b}\right|\cos \theta \\ =2-2\cos \theta \\ \Rightarrow \cos \theta =\frac{2-{\left|\hat{a}-\hat{b}\right|}^2}{2}...\left(2\right) \\ \left(\mathrm{i}\right)\text{Now,} \\ \cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}} \\ =\sqrt{\frac{1+\frac{{\left|\hat{a}+\hat{b}\right|}^2-2}{2}}{2}}\left[\text{From}\left(\text{1}\right)\right] \\ =\sqrt{\frac{2+{\left|\hat{a}+\hat{b}\right|}^2-2}{4}} \\ =\sqrt{\frac{{\left|\hat{a}+\hat{b}\right|}^2}{4}} \\ =\frac{1}{2}\left|\hat{a}+\hat{b}\right| \\ \left(\mathrm{ii}\right)\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ =\sqrt{\frac{1-\frac{2-{\left|\hat{a}-\hat{b}\right|}^2}{2}}{2}}[\text{From (2)]} \\ =\sqrt{\frac{2+{\left|\hat{a}-\hat{b}\right|}^2-2}{4}} \\ =\sqrt{\frac{{\left|\hat{a}-\hat{b}\right|}^2}{4}} \\ =\frac{1}{2}\left|\hat{a}-\hat{b}\right| \\ \text{Now,} \\ \tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{\frac{1}{2}\left|\hat{a}-\hat{b}\right|}{\frac{1}{2}\left|\hat{a}+\hat{b}\right|}=\frac{\left|\hat{a}-\hat{b}\right|}{\left|\hat{a}+\hat{b}\right|} \end{gathered}$$