Polchinio3es

2022-03-24

Formula for $\mathrm{sin}\left\{\frac{\theta}{3}\right\}$

Using$\mathrm{sin}\left\{3\theta \right\}=3\mathrm{sin}\left\{\theta \right\}-4{\mathrm{sin}}^{3}\left\{\theta \right\}$ ,we may calculate $\mathrm{sin}\left\{\frac{\theta}{3}\right\}$

But the formula for root of a cubic equation is very complicated can you help me to find a formula for$\mathrm{sin}\left\{\frac{\theta}{3}\right\}$ as it will help to compute $\mathrm{sin}20}^{\circ},{\mathrm{sin}10}^{\circ$ etc. which will be useful.

Using

But the formula for root of a cubic equation is very complicated can you help me to find a formula for

Korbin Ochoa

Beginner2022-03-25Added 11 answers

Using Cardano's method to solve

$3s-4{s}^{3}=\mathrm{sin}t$

leads to finding the cube roots of

$\pm \sqrt{1-{\mathrm{sin}}^{2}t}+i\mathrm{sin}t=\pm \mathrm{cos}t+i\mathrm{sin}t=\pm {e}^{\mp \text{it}}$

These are non-real number in general, so you end up using polar form, and so go back to square one.

leads to finding the cube roots of

These are non-real number in general, so you end up using polar form, and so go back to square one.

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?

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