Question

Evaluate:

${\left(\frac{1+\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}}{1+\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}}\right)}^6$

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

Answer 1

Answer: (B)

$-1$

Let, $1+cos\frac{\pi }{6}=2co{s}^2\frac{\pi }{12}$ and ..........(1)

$sin\frac{\pi }{6}=2sin\frac{\pi }{12}cos\frac{\pi }{12}$.......(2)

GIven: ${\left(\frac{1+cos\frac{\Pi }{6}-isin\frac{\Pi }{6}}{1+cos\frac{\Pi }{6}+isin\frac{\Pi }{6}}\right)}^6$

$={\left(\frac{2co{s}^2\left(\pi /12\right)-i2sin\left(\pi /12\right)cos\left(\pi /12\right)}{2co{s}^2\left(\pi /12\right)+i2sin\left(\pi /12\right)cos\left(\pi /12\right)}\right)}^6$

$={\left(\frac{2cos\left(\pi /12\right)\left[cos\right(\pi /12)-isin(\pi /12\left)\right]}{2cos\left(\pi /12\right)\left[cos\right(\pi /12)+isin(\pi /12\left)\right]}\right)}^6$

$={\left(\frac{e^{-i\left(\pi /12\right)}}{e^{i\left(\pi /12\right)}}\right)}^6=(e^{-i\left(\pi /6\right)}{)}^6=e^{-i\pi }=cos\pi -isin\pi =-1.$