Question

If the value of $\sqrt{1+\cos \alpha }+\sqrt{1+\cos 2\alpha }+\sqrt{1+\cos 3\alpha }+\cdots +\text{to}n\text{terms}$ is $k\frac{\sin \frac{n\alpha }{4}}{\sin \frac{\alpha }{4}}\cos \left\{\right(n+1\left)\frac{\alpha }{4}\right\},$ then the value of $k^4$ is
(where $0<n\alpha <\pi /2,n\in N$)

Answer 1

$\sqrt{1+\cos \alpha }+\sqrt{1+\cos 2\alpha }+\sqrt{1+\cos 3\alpha }+\cdots +\text{to}n\text{terms}=\sqrt{2{\cos }^2\frac{\alpha }{2}}+\sqrt{2{\cos }^2\alpha }+\sqrt{2{\cos }^2\frac{3\alpha }{2}}+\cdots +\text{to}n\text{terms}=\sqrt{2}[\cos \frac{\alpha }{2}+\cos \alpha +\cos \frac{3\alpha }{2}+\cdots +\text{to}n\text{terms}][\because 0<n\alpha <\pi /2]=\sqrt{2}\frac{\sin \frac{n\alpha }{4}}{\sin \frac{\alpha }{4}}\cos \{\frac{\alpha }{2}+(n-1\left)\frac{\alpha }{4}\right\}=\sqrt{2}\frac{\sin \frac{n\alpha }{4}}{\sin \frac{\alpha }{4}}\cos \left\{\right(n+1\left)\frac{\alpha }{4}\right\}\Rightarrow k=\sqrt{2}\therefore k^4=4$