Solving problem with Double Angle Formula

$ \displaystyle \text{Prove}\ \text{that}\ \ \ \frac{1}{2}\sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=\cos {{72}^{\circ }}.$

$ \displaystyle \text{Solution}$

$ \displaystyle \ \ \ \cos {{72}^{\circ }}=2{{\cos }^{2}}{{36}^{\circ }}-1\ \ \ ........(1)$

$ \displaystyle \ \ \ \cos {{144}^{\circ }}=2{{\cos }^{2}}{{72}^{\circ }}-1$ 

$ \displaystyle \ \ \ -\cos {{36}^{\circ }}=2{{\cos }^{2}}{{72}^{\circ }}-1$

$ \displaystyle \ \ \ \ \ \cos {{36}^{\circ }}=1-2{{\cos }^{2}}{{72}^{\circ }}\ \ .......(2)$

$ \displaystyle (1)+(2)\Rightarrow \ \ \ \cos {{72}^{\circ }}+\cos {{36}^{\circ }}=2\ (\ {{\cos }^{2}}{{36}^{\circ }}-{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cos {{36}^{\circ }}-\cos {{72}^{\circ }}=\frac{1}{2}$

$ \displaystyle (2)-(1)\Rightarrow \ \ \ \cos {{36}^{\circ }}-\cos {{72}^{\circ }}=2-2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}=2-2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})=\frac{3}{2}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }}=\frac{3}{4}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1-{{\sin }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }}=\frac{3}{4}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\sin }^{2}}{{36}^{\circ }}-{{\cos }^{2}}{{72}^{\circ }}=\frac{1}{4}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{\sin }^{2}}{{36}^{\circ }}-4{{\cos }^{2}}{{72}^{\circ }}=1$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{\sin }^{2}}{{36}^{\circ }}-1=4{{\cos }^{2}}{{72}^{\circ }}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=2\cos {{72}^{\circ }}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \frac{1}{2}\sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=\cos {{72}^{\circ }}$