# 3 arccos(x) = arccos(4x$$^{3}$$ - 3x)

We will learn how to prove the property of the inverse trigonometric function 3 arccos(x) = arccos(4x$$^{3}$$ - 3x) or, 3 cos$$^{-1}$$ x = cos$$^{-1}$$ (4x$$^{3}$$ - 3x)

Proof:

Let, cos$$^{-1}$$ x = θ

Therefore, cos θ = x

Now we know that, sin 3θ = 4 cos$$^{3}$$ θ - 3 cos θ

⇒ cos 3θ = 4x$$^{3}$$ - 3x

Therefore, 3θ = cos$$^{-1}$$ (4x$$^{3}$$ - 3x)

⇒ 3 cos$$^{-1}$$ x = cos$$^{-1}$$ (4x$$^{3}$$ - 3x)

or, 3 arccos(x) = arccos(4x$$^{3}$$ - 3x).           Proved.

Inverse Trigonometric Functions

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