I need help writing this in the trigonometric language $$ \frac{\cos(\alpha)-i\sin(\alpha)}{\sin\left(\alpha-\frac{\pi}{2}\right)+i\sin(π-\alpha)}. $$ My biggest doubt is simplifying $\cos(\alpha)-i\sin(\alpha)$. I have been told that I can't write it like $\cos(-\alpha)$. I don't know why. Anyone know how to help?


Note that $$ \sin(\alpha-\pi/2)=-\sin(\pi/2-\alpha)=-\cos\alpha $$ and $\sin(\pi-\alpha)=\sin\alpha$, so the fraction is $$ \frac{\cos\alpha-i\sin\alpha}{-\cos\alpha+i\sin\alpha} $$

If you're compelled to use $\operatorname{cis}$, you can observe that $$ \sin(\alpha-\pi/2)=\sin(\pi/2+\alpha-\pi)=\cos(\pi-\alpha) $$ so the fraction is $$ \frac{\operatorname{cis}(-\alpha)}{\operatorname{cis}(\pi-\alpha)} =\operatorname{cis}(-\alpha-\pi+\alpha)=\operatorname{cis}(-\pi)= \operatorname{cis}(-\pi+2\pi)=\operatorname{cis}\pi=-1 $$

  • $\begingroup$ Thank you, I have understood that part. But I have to simplify the whole fraction. I think the final result is cis(0π). $\endgroup$ – Sarah Jones Mar 18 '17 at 21:25
  • $\begingroup$ @SarahJones I added the argument with cis, but it's a sledgehammer. $\endgroup$ – egreg Mar 18 '17 at 21:55

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