Question

Let $f_n\left(\theta \right)=\sum _{r=0}^n\frac{1}{4^r}{\sin }^4\left(2^r\theta \right)$, then which of the following alternative (s) is/are correct?

• A. $f_2\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$
• B. $f_3\left(\frac{\pi }{8}\right)=\frac{2+\sqrt{2}}{4}$
• C. $f_4\left(\frac{3\pi }{2}\right)=1$
• D. $f_5\left(\pi \right)=0$

Answer 1

Answer: (C), (D)

The correct options are
C $f_4\left(\frac{3\pi }{2}\right)=1$
D $f_5\left(\pi \right)=0$

$f_2\left(\frac{\pi }{4}\right)={\sin }^4\left(\frac{\pi }{4}\right)+\frac{1}{4}{\sin }^4(2\times \frac{\pi }{4})+\frac{1}{4^2}{\sin }^4(4\times \frac{\pi }{4})$

$=\frac{1}{4}+\frac{1}{4}+0=\frac{1}{2}$

$f_3\left(\frac{\pi }{8}\right)={\sin }^4\left(\frac{\pi }{8}\right)+\frac{1}{4}{\sin }^4(2\times \frac{\pi }{8})+\frac{1}{4^2}{\sin }^4(4\times \frac{\pi }{8})+\frac{1}{4^3}{\sin }^4(8\times \frac{\pi }{8})$

$=\frac{3-2\sqrt{2}}{8}+\frac{1}{16}+\frac{1}{16}=\frac{2-\sqrt{2}}{4}$

$f_4\left(\frac{3\pi }{2}\right)={\sin }^4\left(\frac{3\pi }{2}\right)+\frac{1}{4}{\sin }^4\left(3\pi \right)+\frac{1}{4^2}{\sin }^4\left(6\pi \right)+\frac{1}{4^3}{\sin }^4\left(12\pi \right)+\frac{1}{4^4}{\sin }^4\left(24\pi \right)$

$=1+0+0+0=1$

$f_5\left(\pi \right)=0$