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If $b=\mathrm{sin}\left({40}^{\circ }+\alpha \right)$ and ${0}^{\circ }<\alpha <{45}^{\circ }$, compute $\mathrm{cos}\left({70}^{\circ }+\alpha \right)$ in terms of b
I wrote $\mathrm{cos}\left({70}^{\circ }+\alpha \right)=\mathrm{cos}\left({30}^{\circ }+{40}^{\circ }+\alpha \right)=\frac{\sqrt{3}}{2}\mathrm{cos}\left({40}^{\circ }+\alpha \right)-\frac{1}{2}b,$, this didn't work. Then expanded $\mathrm{sin}\left({40}^{\circ }+\alpha \right)$ and $\mathrm{cos}\left({70}^{\circ }+\alpha \right)$ in the hope of getting a hint. Still and all, I couldn't get the answer up above.
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Allison Pena
$\mathrm{cos}\left({70}^{\circ }+\alpha \right)=\mathrm{cos}\left({40}^{\circ }+\alpha +{30}^{\circ }\right)=$
$\mathrm{cos}\left({40}^{\circ }+\alpha \right)\mathrm{cos}\left({30}^{\circ }\right)-\mathrm{sin}\left({40}^{\circ }+\alpha \right)\mathrm{sin}\left({30}^{\circ }\right)=$
$\sqrt{1-\mathrm{sin}\left({40}^{\circ }+\alpha {\right)}^{2}}\cdot \frac{\sqrt{3}}{2}-\frac{b}{2}$
$=\sqrt{1-{b}^{2}}\cdot \frac{\sqrt{3}}{2}-\frac{b}{2}$