Question

Any complex number in the polar form can be expressed in Euler's form as $\cos \theta +i\sin \theta =e^{i\theta }$. This form of the complex number is useful in finding the sum of series $\sum _{r=0}^n{}^n{C}_r(\cos \theta +i\sin \theta {)}^r$.
$\sum _{r=0}^n{}^n{C}_r(\cos r\theta +i\sin r\theta )=\sum _{r=0}^n{}^n{C}_r{e}^{ir\theta }=\sum _{r=0}^n{}^n{C}_r(e^{i\theta }{)}^r=(1+e^{i\theta }{)}^n$
Also, we know that the sum of binomial series does not change if $r$ is replaced by $n-r$. Using these facts, answer the following questions.

If $f\left(x\right)=\frac{\sum _{r=0}^{50}{}^{50}{C}_r\sin 2rx}{\sum _{r=0}^{50}{}^{50}{C}_r\cos 2rx}$, then the value of $f\left(\frac{\pi }{8}\right)$ is equal to

• A. $1$
• B. $-1$
• C. irrational value
• D. $0$

Answer 1

The correct option is A $1$

$\frac{\sum _{r=0}^{50}{}^{50}{C}_r\sin 2rx}{\sum _{r=0}^{50}{}^{50}{C}_r\cos 2rx}$
$=\frac{\sum _{r=0}^{50}{}^{50}{C}_{50-r}\sin 2(50-r)x}{\sum _{r=0}^{50}{}^{50}{C}_{50-r}\cos 2(50-r)x}$

$=\frac{\sum _{r=0}^{50}{}^{50}{C}_r[\sin 2rx+\sin 2(50-r\left)x\right]}{\sum _{r=0}^{50}{}^{50}{C}_r[\cos 2rx+\cos 2(50-r\left)x\right]}$
$(\because \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d})$

$=\frac{\sum _{r=0}^{50}{}^{50}{C}_r\times 2\sin \left(50x\right)\cos (2r-50)x}{\sum _{r=0}^{50}{}^{50}{C}_r\times 2\cos \left(50x\right)\cos (2r-50)x}$

$=\tan \left(50x\right)$

$\therefore f\left(\frac{\pi }{8}\right)=\tan \left(\frac{25\pi }{4}\right)=\tan (6\pi +\frac{\pi }{4})=1$