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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.

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

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

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

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