Show that $\sin x-i\cos x=\operatorname{cis}(x-\pi/2)$

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Show that $\sin\theta-i\cos\theta=\operatorname{cis}(\theta-\frac{\pi}{2})$

I've previously shown that: $\sin\theta+i\cos\theta=\operatorname{cis}(\frac{\pi}{2}-\theta)$, and that $\cos\theta-i\sin\theta=\operatorname{cis}(-\theta)$. So I've tried using that in my working out, but can't seem to work this one out.

I am assuming that it wants me to use trig addition formulae and De Moivre's Theorem possibly (because that is the general theme of this section of the book). No working out that's too abstract or advanced I'd presume.

I have tried symmetry properties, and negative angles as well.

I am sure I can prove it by using a subtraction formula, but that would be me assuming the answer (circular logic maybe). I need to start from the LHS and then work to get the RHS.

You can write $$\operatorname{cis}\left(\theta-\frac{\pi}{2}\right)= \frac{\operatorname{cis}\theta}{\operatorname{cis}\frac{\pi}{2}}$$ Since $\operatorname{cis}\frac{\pi}{2}=i$, you get $$\operatorname{cis}\left(\theta-\frac{\pi}{2}\right)= i^{-1}\operatorname{cis}\theta=i^{-1}(\cos\theta+i\sin\theta)= -i\cos\theta+\sin\theta$$ because $i^{-1}=-i$.

• Thank you for that. It ties in very well with the more recent sections. – Simplex1 Aug 3 '18 at 11:57

Well, we have:

$$\exp\left(\theta i\right)=\cos\left(\theta\right)+\sin\left(\theta\right)i\tag1$$

So, when we have:

$$\exp\left(\left(\theta-\frac{\pi}{2}\right)\cdot i\right)=\cos\left(\theta-\frac{\pi}{2}\right)+\sin\left(\theta-\frac{\pi}{2}\right)i\tag2$$

And we know that:

• $$\cos\left(\theta-\frac{\pi}{2}\right)=\sin\left(\theta\right)\tag3$$
• $$\sin\left(\theta-\frac{\pi}{2}\right)=-\cos\left(\theta\right)\tag4$$