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.


















11 and 12 Grade Math

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