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How do I evaluate $ sin(20)$ exactly? [in degrees]

I derived the relationship between $sin(x) $ and $sin(3x)$ where $x = sin(x)$ and $ y = sin(3x)$

Now I am interested in subsituting $sin(60) $ and moving along but I am not sure which formula will result in me getting a real solution.

I am curious how to de-nest this mess into something cleaner. I would really like to get rid of all the imaginary numbers but if that is not possible I still feel that this can definitely be de-nested into a simpler looking form even if the radical count does not go down.

I have tried a couple attempts at substituting stuff back in but my answer seems to change whenever I move thing in and out of the cube root

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Sorry but, what is the link between the title and your question? – L. F. Sep 3 '13 at 1:10
My mistake... I was going to ask something else but I forgot to change the title – frogeyedpeas Sep 3 '13 at 1:12
Title has been fixed – frogeyedpeas Sep 3 '13 at 1:14
@frogeyedpeas - with over 1000 rep points, you should know to use LaTeX formatting so that others don't have to edit your post for you. – nbubis Sep 3 '13 at 1:14
This doesn't look like something that would need it. The only thing that will change is sin(60) becomes $\sin(60)$ – frogeyedpeas Sep 3 '13 at 1:15
up vote 1 down vote accepted

So the solution to this equation corresponding to the actual value of $\sin(20˚)$ is given by WA as the second solution in your second link as $$ x = -\frac{1-i \sqrt3}{2\cdot2^{2/3}} \left(-\sqrt{3}+i\right)^{1/3}-\frac14 \left(\frac12 (-\sqrt{3}+i)\right)^{1/3} (1+i \sqrt3) $$ We can rewrite this as $$ x=\left[-\frac{1-i \sqrt3}{2\cdot2^{1/3}} -\frac14 (1+i \sqrt3)\right]\left(\frac12 (-\sqrt{3}+i)\right)^{1/3} $$ This in turn becomes $$ x=\left[\left(-\frac14 -\frac1{2 \times2^{1/3} }\right) + \left(\frac1{2 \times2^{1/3} }-\frac14 \right)i\sqrt3\right] \left(\frac12 (-\sqrt{3}+i)\right)^{1/3} $$ Maybe that will be a little easier to work with

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You can't get an expression for this using real radicals: see

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