Method 1double angle identity for cosine: \(\cos 2\alpha = 2{\cos ^2}\alpha - 1\)substituting \(\theta \) for \(2\alpha \) gives: \(\cos \theta = 2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1\;\)from the cosine rule: \(\cos \theta = \frac{{{x^2} + {x^2} - {a^2}}}{{2xx}} = \frac{{2{x^2} - {a^2}}}{{2{x^2}}}\)substituting \(2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1\) for \(\cos \theta \), gives:\(2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1 = \frac{{2{x^2} - {a^2}}}{{2{x^2}}}\)\(2{\cos ^2}\left( {\frac{\theta }{2}} \right) = \frac{{2{x^2} - {a^2}}}{{2{x^2}}} + 1\) \( = \frac{{2{x^2} - {a^2}}}{{2{x^2}}} + \frac{{2{x^2}}}{{2{x^2}}}\) \( = \frac{{4{x^2} - {a^2}}}{{2{x^2}}}\) \( = \frac{{4{x^2}}}{{2{x^2}}} - \frac{{{a^2}}}{{2{x^2}}}\) \( = 2 - \frac{{{a^2}}}{{2{x^2}}}\) \( = 1 - \frac{{{a^2}}}{{4{x^2}}}\)\(\cos \left( {\frac{\theta }{2}} \right) = \sqrt {1 - \frac{{{a^2}}}{{4{x^2}}}} \) Q.E.D.Method 2Consider the shaded right triangle shown in figure.\(h = \sqrt {{x^2} - {{\left( {\frac{a}{2}} \right)}^2}} \)