If $0<\theta <180$, then $\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2(1+cos\theta)}}}}$=? For n number of 2s

Question

OPTIONS

A)$2\cos(\frac{\theta}{2^n})$

B)$2\cos(\frac{\theta}{2^{n-1}})$

C)$2\cos\frac{\theta}{2^{n+1}}$

D) none of these

In case of $2(1+cos\theta)$ It can be written $4cos^2\frac{\theta}{2}$.

It’s square will be $2cos\frac{\theta}{2}$

So as we go on multiply 2 and taking its square root, I don’t see why there should be an ‘n’ inside the cos function. Yet there is, and I am wrong, so how should I solve it?

Answer 1

Let $(u_n)$ the sequence defined by $u_0=2\cos{\theta}$ and $u_{n+1}=\sqrt{u_n+2}$ it is quite easy to prove by induction that $u_n=2\cos\left(\dfrac{\theta}{2^n}\right)$

Answer 2

If $n=1$, then you did all the work: $$\sqrt{2(1+\cos\theta)}=\sqrt{4\cos^2(\theta/2)}=2|\cos(\theta/2)|=2\cos(\theta/2)$$ because $0\leq\theta/2\leq\pi/2$, so the cosine of $\theta/2$ is nonnegative.

Now we use the above to get the formula for $n=2$: $$\sqrt{2+\sqrt{2(1+\cos\theta)}}=\sqrt{2+2\cos(\theta/2)}=\sqrt{2(1+\cos(\theta/2)})$$ This is very similar to the case $n=1$! Just use the $\theta/2$ in place of $\theta$: $$\sqrt{2+\sqrt{2(1+\cos\theta)}}=2\cos(\theta/4)$$ Keep repeating this and get $2\cos(\theta/2^n)$ for general $n$.