Question

If $4cos \left(36\right)^{o}+cot\left(7 \left(\frac{1}{2}\right)^{^\circ }\right)=\sqrt{n_{1}}+\sqrt{n_{2}}+\sqrt{n_{3}}$ $+\sqrt{n_{4}}+\sqrt{n_{5}}+\sqrt{n_{6}}$ , then the value of $\left(\frac{\displaystyle \sum _{i = 1}^{6} n_{i}^{2}}{10}\right)$ is equal to

Answer 1

$4cos \left(36\right)^{o}=4\left(\frac{\sqrt{5} + 1}{4}\right)=\sqrt{5}+\sqrt{1}$
$cot 7\left(\frac{1}{2}\right)^{o}=\frac{1 + cos ⁡ \left(15\right)^{^\circ }}{sin ⁡ \left(15\right)^{^\circ }}=\frac{1 + \left(\frac{\sqrt{3} + 1}{2 \sqrt{2}}\right)}{\left(\frac{\sqrt{3} - 1}{2 \sqrt{2}}\right)}=\frac{2 \sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}$
$=\frac{\left(2 \sqrt{2} + \sqrt{3} + 1\right) \left(\sqrt{3} + 1\right)}{\left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right)}=\frac{2 \sqrt{6} + 2 \sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{2}$
$=\frac{2 \sqrt{6} + 2 \sqrt{3} + 2 \sqrt{2} + 4}{2}=\sqrt{6}+\sqrt{3}+\sqrt{2}+2=\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}$
$\Rightarrow 4cos \left(36\right)^{o}+cot\left(7 \left(\frac{1}{2}\right)^{o}\right)=\sqrt{5}+\sqrt{1}+\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}$
$=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}$
$=\sqrt{n_{1}}+\sqrt{n_{2}}+\sqrt{n_{3}}+\sqrt{n_{4}}+\sqrt{n_{5}}+\sqrt{n_{6}}$
$\Rightarrow \displaystyle \sum _{i = 1}^{6} n_{i}^{2} = 1^{2} + 2^{2} + 3^{3} + 4^{2} + 5^{2} + 6^{2}$