I am trying to solve the problem: "Multiply $[5\operatorname{cis}(-70^\circ)][3\operatorname{cis}(280^\circ)]$ and express the answer in rectangular coordinates. Give an exact answer." I have my prior answer available, which was marked correct, but not my work.

Presently, I expanded the expression to

$$15\cos(-70)\cos(280) + 15i\cos(-70)\sin(280) + 15i\sin(-70)\cos(280) - 15\sin(-70)\sin(280)$$

but I'm not sure how to find exact fractional solutions to the individual trig expressions, rather than decimal approximations. Wolfram Alpha, by contrast, returns $\sin(\pi/18)$ (another trig expression) as the result of $\cos(280)$, rather than a fractional/radical answer.

I also noted that $\operatorname{cis}(x) = e^{ix}$, but I'm told this equation doesn't work when $x$ is expressed in degrees. It seems like a trivial workaround to convert degrees to radians, i.e.,

$$\operatorname{cis}(280) \implies \operatorname{cis}\left(\frac{14\pi}{9}\right)$$

apply the equation

$$\operatorname{cis}\left(\frac{14\pi}{9}\right) \implies e^{i(14\pi)/9}$$

and then convert back to degrees

$$e^{i(14\pi)/9} \implies e^{280i}$$

but this yields the same answer as using the equation on the angle expressed in degrees in the first place, so it's unlikely to be legal.

What is the best way to solve this?

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 20 '18 at 9:47

\begin{align}5\operatorname{cis}(-70^\circ) \cdot 3 \operatorname{cis}(280^\circ)&=(5\cdot3) \operatorname{cis}(280^\circ-70^\circ) \\ &=15 \operatorname{cis}(210^\circ)\\ &= 15(\cos(210^\circ)+i\sin(210^\circ))\\ &=-15(\cos(30^\circ)+i\sin(30^\circ))\\ &= -\frac{15(\sqrt3+i)}2\end{align}

  • $\begingroup$ What identity did you use for the first line? Did you expand and use these wikimedia.org/api/rest_v1/media/math/render/svg/… or can you also just substitute sin(a + b) for sin(a) sin(b) and cos(a + b) for cos(a) cos(b) respectively? $\endgroup$ – user10478 Apr 20 '18 at 14:21
  • $\begingroup$ $\operatorname{cis}(a)\cdot\operatorname{cis}(b) = \exp(ia) \exp(ib) = \exp(i(a+b)=\operatorname{cis}(a+b)$. $\endgroup$ – Siong Thye Goh Apr 20 '18 at 16:50

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