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Given that $\cos{160^{\circ}} = -q$, express $\cos70^{\circ}$ in terms of $q$.

No example in the book, don't know how to do it??

I need a complete explanation.

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$$\cos(160^{\circ})=-q$$ $$\cos(180^{\circ}-20^{\circ})=-q$$ $$-\cos(20^{\circ})=-q$$ Because $\cos(180^{\circ}-x)=-\cos(x)$ $$\cos(20^{\circ})=q$$

$$\cos(70^{\circ})=\cos(90^{\circ}-20^{\circ})=\sin(20^{\circ})=\sqrt{1-q^2}$$

First because $\cos(90^{\circ}-x)=\sin(x)$ and second $\cos^2(x)+\sin^2(x)=1$ and the fact $20 ^{\circ}$ is in the first quadrant (and then is positive)

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Hint: $\cos 70^\circ=\cos\, (160^\circ - 90^\circ)=\sin 160^\circ$.

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