# A proving question based on DeMoivre's theorem.

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.

• – lab bhattacharjee Oct 21 '18 at 6:23
• The correct spelling is deMoivre ..... or de Moivre. – DanielWainfleet Oct 21 '18 at 8:04

## 3 Answers

\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}

Hope it helps...just expand 1 in terms of sin and cos • Did it help you? – user579252 Oct 21 '18 at 4:00
• Are you sure that $e^{i\pi /8} =-1$? – Mohammad Riazi-Kermani Oct 21 '18 at 4:51
• @MohammadRiazi-Kermani Actually this question is wrong sir ; RHS will have sinpi/8 + i cospi/8 – user579252 Oct 21 '18 at 5:02
• If the question is wrong then how did you get that answer? – Key Flex Oct 21 '18 at 5:17

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.