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
11 and 12 Grade Math
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