# Trigonometry simplification with $\cos(\alpha)$

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?

## 1 Answer

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

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