Question

show that 3cos _1x=cos_1(4cube _ 3x)xE(1/2,1)​

Answer 1

Answer:

Let x=cosθ⇒cos−1x=θ.

Thus we have,

R.H.S.=cos−1(4x3−3x)

=cos−1(4cos3θ−3cosθ)

=cos−1(cos3θ)

=3θ=3cos−1x=L.H.S.

Answer 2

Answer:

We have to prove that 3cos−1x=cos−1(4x3−3x)

First consider RHS,  cos−1(4x3−3x)

Let us take x=cosθ, we know that cos(3θ)=4cos3θ−3cosθ

⟹cos−1(4x3−3x)=cos−1(4cos3θ−3cosθ)=cos−1(cos(3θ))=3θ

∵θ=cos−1x, ⟹cos−1(4x3−3x)=3cos−1x

∴LHS=RHS

0≤3cos−1x≤π

0≤cos1x≤π/3

∴xϵ[21,1]

Hence proved.

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