Evaluate-1-cos-pi-10-isin-pi-10-1-cos-pi-10-isin-pi-10-15-

Question Number 144697 by ZiYangLee last updated on 28/Jun/21

Evaluate {(\frac{1 + \cos \frac{π}{10} - i \sin \frac{π}{10}}{1 + \cos \frac{π}{10} + i \sin \frac{π}{10}})}^{15} .

Answered by mathmax by abdo last updated on 28/Jun/21

1 + \cos (\frac{π}{10}) + isin (\frac{π}{10}) = {2 \cos}^{2} (\frac{π}{20}) + 2 isin (\frac{π}{20}) \cos (\frac{π}{20})

= 2 \cos (\frac{π}{20}) e^{\frac{i π}{20}} and 1 + \cos (\frac{π}{10}) - isin (\frac{π}{10}) = 2 \cos (\frac{π}{20}) e^{- \frac{i π}{20}} \Rightarrow

A = \frac{1 + \cos (\frac{π}{10}) - isin (\frac{π}{10})}{1 + \cos (\frac{π}{10}) + isin (\frac{π}{10})} = \frac{e^{- \frac{i π}{20}}}{e^{\frac{i π}{20}}} = e^{- \frac{i π}{10}} \Rightarrow

A^{15} = e^{- \frac{i 15 π}{10}} = e^{i (\frac{20 - 5) π}{10})} = e^{2 i π} . e^{- \frac{i π}{2}} = - i

Answered by som(math1967) last updated on 28/Jun/21

\frac{1 + c o s θ - i s i n θ}{1 + c o s θ + i s i n θ} [θ = \frac{π}{10}]

= \frac{{(1 + c o s θ - i s i n θ)}^{2}}{{(1 + c o s θ)}^{2} - i^{2} {s i n}^{2} θ}

= \frac{{(1 + c o s θ)}^{2} - 2 i s i n θ (1 + c o s θ) + i^{2} {s i n}^{2} θ}{1 + 2 c o s θ + {c o s}^{2} θ + {s i n}^{2} θ}

= \frac{{(1 + c o s θ)}^{2} - 2 i s i n θ (1 + c o s θ) - (1 - {c o s}^{2} θ)}{2 (1 + c o s θ)}

= \frac{(1 + c o s θ) (1 + c o s θ - 2 i s i n θ - 1 + c o s θ)}{2 (1 + c o s θ)}

= \frac{2 (c o s θ - i s i n θ)}{2} = (c o s θ - i s i n θ)

∴ (\frac{1 + c o s \frac{π}{10} - i s i n \frac{π}{10}}{1 + c o s \frac{π}{10} + i s i n \frac{π}{10}}) = (c o s \frac{π}{10} - i s i n \frac{π}{10})

{(\frac{1 + c o s \frac{π}{10} - i s i n \frac{π}{10}}{1 + c o s \frac{π}{10} + i s i n \frac{π}{10}})}^{15}

= {(c o s \frac{π}{10} - i s i n \frac{π}{10})}^{15} = c o s \frac{3 π}{2} - i s i n \frac{3 π}{2}

= 0 - i (- 1) = i a n s