A proving question based on DeMoivre's theorem.

Question

Prove that$$\frac{1+\sin(1/8)π+i \cos(1/8)π}{1+\sin(1/8)π–i \cos(1/8)π} =\; –1$$ I tried to solve this by converting it into $e^{ik\alpha}$ but could not rationalize it please help me out.

Answer 1

Hope it helps...just expand 1 in terms of sin and cos

Answer 2

The given equality is not true.

Upon cross multiplication of $$\frac{1+\sin(1/8)π+i \cos(1/8)π}{1+\sin(1/8)π–i \cos(1/8)π} =\; –1$$

We get $$1+\sin(1/8)π+i \cos(1/8)π= -1-\sin(1/8)π+i \cos(1/8)π$$

Which is equivalent to $$1+\sin(1/8)π= -1-\sin(1/8)π$$ or $$\sin(1/8)π=-1$$ which is obviously false.

Answer 3

$$ \begin{align} \frac{1+\sin(\pi/8)+i\cos(\pi/8)}{1+\sin(\pi/8)-i\cos(\pi/8)} &=\frac{1+ie^{-i\pi/8}}{1-ie^{i\pi/8}}\frac{ie^{-i\pi/8}}{ie^{-i\pi/8}}\\ &=\frac{1+ie^{-i\pi/8}}{1+ie^{-i\pi/8}}ie^{-i\pi/8}\\[7pt] &=ie^{-i\pi/8}\\[12pt] &=e^{i3\pi/8} \end{align} $$