Trigonometry & Inverse Trigonometry Question - None

Question

If $\cos^{-1}(2x^2-1)=2\pi-2\cos^{-1}x$ , then:

Accepted Answer

Step 1: Let $\theta=\cos^{-1}x\in[0,\pi]$ . Equation: $\cos^{-1}(\cos 2\theta)=2\pi-2\theta$ .

Step 2: $\cos^{-1}(\cos\alpha)=2\pi-\alpha$ is valid for $\alpha\in[\pi,2\pi]$ .

Step 3: Need $2\theta\in[\pi,2\pi]\implies\theta\in[\pi/2,\pi]\implies\cos^{-1}x\in[\pi/2,\pi]\implies x\in[-1,0]$ .

Answer: (A) $x\in[-1,0]$

Trap: Standard property:

\cos^{-1}(2x^2-1)=2\cos^{-1}x

for

x\in[0,1]

and

2\pi-2\cos^{-1}x

for

x\in[-1,0]

.

Key Concept: Branch-tracking in double-angle cos⁻¹ identities

Question: