0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$ \frac{\operatorname{cis}(-\alpha)}{\operatorname{cis}(\pi-\alpha)} =\operatorname{cis}(-\alpha-\pi+\alpha)=\operatorname{cis}(-\pi)= \operatorname{cis}(-\pi+2\pi)=\operatorname{cis}\pi=-1 $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.