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I have this trig problem where it asks to find $ \sin(10^{\circ})$, given $\cos(40^{\circ})=a$. I tried the write the $\sin(10^{\circ})$ as $\sin(40^{\circ}-30^{\circ})$ to use the sum formula but i didn't get anywhere near to the options given in the test:

a)$\sin(10^{\circ})=2a$

b)$\sin(10^{\circ})=2a^2$

c)$\sin(10^{\circ})=2a-1$

d)$\sin(10^{\circ})=2a^2+1$

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2 Answers

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We have used here two facts $$\sin { \left( \alpha \right) =\cos { \left( 90^{ \circ }-\alpha \right) } } \\ \cos { \left( 2\alpha \right) =\cos ^{ 2 }{ \alpha -\sin ^{ 2 }{ \alpha } } =\cos ^{ 2 }{ \alpha -\left( 1-\cos ^{ 2 }{ \alpha } \right) =2\cos ^{ 2 }{ \alpha -1 } } } $$

$$\sin { \left( 10^{ \circ } \right) =\cos { \left( 80^{ \circ } \right) =2\cos ^{ 2 }{ \left( 40^{ \circ } \right) } -1 } =2{ a }^{ 2 }-1 } $$

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You are on the right lines

\begin{align*} \sin(10) &= \sin(40-30) \\ &= \sin(40)\cos(30)-\cos(40)\sin(30) \\ &= \frac{\sqrt{3}}{2}\sin(40) - \frac{1}{2}\cos(40) \\ &= \frac{\sqrt{3}}{2}\sqrt{1-\cos^2(40)} - \frac{1}{2}\cos(40) \\ &= \frac{\sqrt{3}}{2}\sqrt{1-a^2} - \frac{1}{2}a \\ \end{align*}

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