<s>[INST] Find $\sin(10^{\circ})$, if $\cos(40^{\circ}) = a$ 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)$\sin(10^{\circ})=2a^2+1$
 [/INST] 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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