Express cos⁡(α3) as a function of cos⁡(α) Give an expression for cos⁡(α3) one can do the following trick: cos⁡(3x)=cos⁡(2x+x)=cos⁡(2x)cos⁡(x)−sin⁡(2x)sin⁡(x) =[cos2(x)−sin2(x)]cos−2sin2(x)cos⁡(x)=4cos3(x)−3cos⁡(x) By defining α=3x, we have 4cos3(α3)−3cos⁡(α3)=cos⁡(α) However, I have not found any way to obtain an expression for cos⁡(α3) solely as a function of cos⁡(α), as one can instead do with cos⁡(α2)=±121+cos⁡(α) Is it simply impossible or is there a way to obtain the desired result?

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Express cos⁡(α3) as a function of cos⁡(α)  Give an expression for cos⁡(α3) one can do the following trick:  cos⁡(3x)=cos⁡(2x+x)=cos⁡(2x)cos⁡(x)−sin⁡(2x)sin⁡(x)  =[cos2(x)−sin2(x)]cos−2sin2(x)cos⁡(x)=4cos3(x)−3cos⁡(x)  By defining α=3x, we have  4cos3(α3)−3cos⁡(α3)=cos⁡(α)  However, I have not found any way to obtain an expression for cos⁡(α3) solely as a function of cos⁡(α), as one can instead do with cos⁡(α2)=±121+cos⁡(α)  Is it simply impossible or is there a way to obtain the desired result?

alkaholikd9 · Accepted Answer

It is indeed impossible. There are closed-form formulas for the roots of a cubic equation. But in the case of three real roots, known as the casus irreductibilis, the solution requires… angle trisection (in fact, the cubic root of a complex), i.e. precisely the problem you are asking. On another hand, expressing coshx3 in terms of coshx is doable: coshx+cosh2x−1=ex ex3=ex3 coshx3=12ex3+1ex3

Joseph Fair · Accepted Answer

Since cos⁡0≠cos⁡2π3  ,cos⁡α3 isn&#039;t a function of cos⁡α.

Express \cos(\frac{\alpha}{3}) as a function of \cos(\alpha) Give an expression for

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