Best answer:
Lots of radicals, if it can be done at all.
sin(30) = sin(20)cos(10) + cos(20)sin(10)
= 2*sin(10)cos(10)cos(10) + cos^2(10)sin(10) - sin^3(10) =>
sin(30) = 3*cos^2(10)*sqrt(1 - cos^2(10)) - (1 - cos^2(10))*sqrt(1 - cos^2(10)) =>
1/2 = sqrt(1 - cos^2(10))*[3*cos^2(10) - 1 + cos^2(10)]
= sqrt(1 -...
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Best answer: Lots of radicals, if it can be done at all.
sin(30) = sin(20)cos(10) + cos(20)sin(10)
= 2*sin(10)cos(10)cos(10) + cos^2(10)sin(10) - sin^3(10) =>
sin(30) = 3*cos^2(10)*sqrt(1 - cos^2(10)) - (1 - cos^2(10))*sqrt(1 - cos^2(10)) =>
1/2 = sqrt(1 - cos^2(10))*[3*cos^2(10) - 1 + cos^2(10)]
= sqrt(1 - cos^2(10))*(4*cos^2(10) - 1).
Hence,
1/sqrt(1 - cos^2(10)) = 8*cos^2(10) - 2 =>
1/(1 - cos^2(10)) = 64*cos^4(10) - 32*cos^2(10) + 4 =>
1 = [64*cos^4(10) - 32*cos^2(10) + 4] * (1 - cos^2(10))
This will give a cubic in [cos^2(10)], and a cubic can be SOLVED by the "cubic formula" which is quite ugly. So you CAN get a fraction with a bunch of radicals in it.
And then of course sin^2(10) = 1 - cos^2(10), so you CAN get a fraction for that too.
7 answers
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19 hours ago