Question

Prove that \(3\cos^{-1}x=\cos^{-1}(4x^3-3x),\ x\in\left[\dfrac12,1\right]\).

Hint:

Substitute \(x=\cos\theta\) and use \(\cos3\theta=4\cos^3\theta-3\cos\theta\); check \(3\theta\) stays within \([0,\pi]\).

Answer 1

Step 1: Understanding the Concept:
The expression \(4x^3-3x\) is exactly the triple-angle formula \(\cos3\theta=4\cos^3\theta-3\cos\theta\) in disguise; substituting \(x=\cos\theta\) should make the identity fall out.

Step 2: Substituting x = cos(theta):
Let \(x=\cos\theta\), so \(\theta=\cos^{-1}x\). Since \(x\in[\tfrac12,1]\), \(\theta\in[0,\tfrac{\pi}{3}]\).

Step 3: Applying the triple angle formula:
\(4x^3-3x=4\cos^3\theta-3\cos\theta=\cos3\theta\).

Step 4: Checking the range for the inverse cosine to apply directly:
Since \(\theta\in[0,\pi/3]\), we get \(3\theta\in[0,\pi]\), which is exactly the principal range of \(\cos^{-1}\), so \(\cos^{-1}(\cos3\theta)=3\theta\) without any adjustment.

Final Answer:
Hence \(\cos^{-1}(4x^3-3x)=3\theta=3\cos^{-1}x\), i.e. \(\boxed{3\cos^{-1}x=\cos^{-1}(4x^3-3x)}\).