Prove: #3cos^-1x=cos^-1(4x^3-3x)#?

2 Answers
May 13, 2018

To prove #3cos^-1x=cos^-1(4x^3-3x)#

Let #cos^-1x=theta#

#=>x=costheta#

Now #LHS=3theta#

#=cos^-1cos(3theta)#

#=cos^-1(4cos^3theta-3costheta)#

#=cos^-1(4x^3-3x)#

May 13, 2018

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# 3 arccos x = arccos (4x^3 -3 x )#

Sometimes trig is less about doing math and more about recognizing math when we see it. Here we recognize #4x^3 -3x# as the cosine triple angle formula, #\cos(3 \theta)# when #x=\cos \theta#.

Factoid: #4x^3-3x# is also called #T_3(x)#, the third Chebyshev Polynomial of the first kind. In general, #\cos(nx) = T_n(\cos x).#

We'll assume #arccos# refers to the principal value. I prefer to call the principal #text{Arc}text{cos}# but that's harder to type.

Enough background. Once we've recognized the triple angle formula the proof is easy.

Proof :

Let #theta = arccos x.#

#x=cos theta#

# cos 3 theta = 4 cos^3 theta - 3 cos theta #

# cos 3 (arccos x) = 4x^3 - 3 x #

# 3 arccos x = arccos(4x^3 - 3x) quad sqrt#