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The expression $$\dfrac{(\text{cis}\ 75^\circ-\text{cis}\ 155^\circ)(1-\cos 8^\circ+i \sin 8 ^\circ)}{2-2\cos 8^\circ}$$ can be written as $r\ \text{cis}\ \theta,$ where $0 \le \theta < 360^\circ$. Find $\theta$ in degrees.

I expanded the numerator to $$\text{cis}\ 75^\circ-\text{cis}\ 75^\circ\cos 8 ^\circ+i \ \text{cis}\ 75^\circ\sin 8^\circ-\text{cis}\ 155^\circ+\text{cis}\ 155^\circ \cos 8^\circ-i \ \text{cis}\ 155^\circ\sin 8^\circ.$$ But I don't think I can simplify anymore. Is my approach a dead end or did I miss some trig identities?

*For those that are unfamiliar with $\text{cis}$, $\text{cis} \ \theta=\cos \theta+i \sin \theta$.

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    It might be clear or if you wrote $\mathrm{cis}$ in exponential form. – 2017-01-14

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Expanding is here not the easiest solution strategy, better combine terms. $$ e^{i·75°}-e^{i·155°}=-e^{i·115°}·2i\sin(40°), $$ and $$ 1-e^{-i·8°}=e^{-i·4°}·2i\sin(4°). $$ and in general $$ e^{ib}-e^{ia}=e^{i\frac{b+a}2}·2i\sin\frac{b-a}2 $$