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#### jacobi

##### Active member

- May 22, 2013

- 58

Find the exact value of \(\displaystyle \sin \frac{\pi}{180}\).

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- Thread starter
- #1

- May 22, 2013

- 58

Find the exact value of \(\displaystyle \sin \frac{\pi}{180}\).

- Aug 30, 2012

- 1,143

Ummmmm....is the argument in radians or degrees? Kinda hard to tell given the fraction.Find the exact value of \(\displaystyle \sin \frac{\pi}{180}\).

-Dan

Just ask the Wolf. HEREFind the exact value of \(\displaystyle \sin \frac{\pi}{180}\).

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- Feb 7, 2012

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Start with the known formula for $\sin 3^\circ$ (you can find it here): $\sin 3^\circ = \frac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr).$ The formula $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$ then tells you that $\sin1^\circ$ is the smaller of the two positive roots of the cubic equation $$4x^3 - 3x + \tfrac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr) = 0,$$ which can be solved exactly, for example by Vieta's method. But don't expect a neat solution.Find the exact value of \(\displaystyle \sin \frac{\pi}{180}\) (in other words, $\color{red}{\sin 1^\circ}$).

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- Jan 29, 2012

- 661

Related with the OP (but more simple): sin 72º.

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- #6

- May 22, 2013

- 58

By convention, if there is no degree symbol in the argument, it is in radians.Ummmmm....is the argument in radians or degrees? Kinda hard to tell given the fraction.

-Dan

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- May 22, 2013

- 58

My method was exactly the same as yours, but I went ahead and solved the cubic. I found that the smallest positive solution can be written as the imaginary part of \(\displaystyle \sqrt[3]{ \frac{a+b}{4}}\), where \(\displaystyle a=\sqrt{8+\sqrt{15}+\sqrt{3}+\sqrt{10-2 \sqrt{5}}}\) and \(\displaystyle b=\sqrt{-8+\sqrt{15}+\sqrt{3}+\sqrt{10-2 \sqrt{5}}}\). Conversely, the cosine of \(\displaystyle 1^\circ\) can be written as the real part of the above.Start with the known formula for $\sin 3^\circ$ (you can find it here): $\sin 3^\circ = \frac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr).$ The formula $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$ then tells you that $\sin1^\circ$ is the smaller of the two positive roots of the cubic equation $$4x^3 - 3x + \tfrac1{16}\Bigl(2(1-\sqrt3)\sqrt{5+\sqrt5} + \sqrt2(\sqrt5-1)(\sqrt3+1) \Bigr) = 0,$$ which can be solved exactly, for example by Vieta's method. But don't expect a neat solution.