Question

Evaluate:

${\left(\frac{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}\right)}^8$

• A. $1$
• B. $-1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $-1$

Let, $1+cos\frac{\pi }{8}=2co{s}^2\frac{\pi }{16}$ and ..........(1)

$sin\frac{\pi }{8}=2sin\frac{\pi }{16}cos\frac{\pi }{16}$.......(2)

And $e^{\pm i\theta }=cos\theta \pm isin\theta$

GIven: ${\left(\frac{1+cos\frac{\Pi }{8}-isin\frac{\Pi }{8}}{1+cos\frac{\Pi }{8}+isin\frac{\Pi }{8}}\right)}^8$

$={\left(\frac{2co{s}^2\left(\pi /16\right)-i2sin\left(\pi /16\right)cos\left(\pi /16\right)}{2co{s}^2\left(\pi /16\right)+i2sin\left(\pi /16\right)cos\left(\pi /16\right)}\right)}^8$

$={\left(\frac{2cos\left(\pi /16\right)\left[cos\right(\pi /16)-isin(\pi /16\left)\right]}{2cos\left(\pi /16\right)\left[cos\right(\pi /16)+isin(\pi /16\left)\right]}\right)}^8$

$={\left(\frac{e^{-i\left(\pi /16\right)}}{e^{i\left(\pi /16\right)}}\right)}^8=(e^{-i\left(\pi /8\right)}{)}^8=e^{-i\pi }=cos\pi -isin\pi =-1.$