Question

Show that $\cot 7{\frac{1}{2}}^{\circ }=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}.$

Answer 1

$\cos {\left(7\frac{1}{2}\right)}^o=\cos {\left(\frac{15}{2}\right)}^o=\frac{\cos {15}^o-1}{2}=\frac{\cos ({45}^o-{30}^o)-1}{2}$

$=\frac{\cos {45}^o\cos {30}^o+\sin {45}^o\sin {30}^o-1}{2}$

using $cos(a-b)=cosa.cosb+sina.sinb$

$=\frac{\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}-1}{2}=\frac{\sqrt{3}+1-2\sqrt{3}}{4\sqrt{2}}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$