How do you simplify #cos((1/3)arccos((1/3)x))#?

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Answer 1 · 2016-10-29T21:42:45

This is already in simplest form, unless you would consider the following simpler:

#cos ((1/3) arccos(x/3))#

#= 1/2((x/3 + i sqrt(1-(x/3)^2))^(1/3) + (x/3 - i sqrt(1-(x/3)^2))^(1/3))#

Expressions of the form #cos(1/3 arccos(1/3 x))# are encountered when solving cubic equations using trigonometric substitution.

For example, consider #x = 3 cos (pi/3) = 3/2#.

Then:

#cos ((1/3) arccos((1/3)x)) = cos(1/3 arccos(1/2)) = cos(pi/9)#

which has no expression in terms of Real #n#th roots.

We can express something in terms of a Complex cube root:

Using de Moivre's Theorem:

#cos ((1/3) arccos(x/3)) + i sin ((1/3) arccos(x/3))#

#= (cos arccos(x/3) + i sin arccos(x/3))^(1/3)#

#= (x/3 + i sqrt(1-(x/3)^2))^(1/3)#

Similarly:

#cos ((1/3) arccos(x/3)) - i sin ((1/3) arccos(x/3))#

#= (cos arccos(x/3) - i sin arccos(x/3))^(1/3)#

#= (x/3 - i sqrt(1-(x/3)^2))^(1/3)#

We can add these two results and halve the sum to find:

#cos ((1/3) arccos(x/3))#

#= 1/2((x/3 + i sqrt(1-(x/3)^2))^(1/3) + (x/3 - i sqrt(1-(x/3)^2))^(1/3))#