Question

Express $y = \cos^{-1}(2x^{2}-1)$ in terms of $\cos^{-1}x$.

• A. $2\cos^{-1}x$
• B. $2\pi - 2\cos^{-1}x$
• C. $2\cos^{-1}(|x|)$
• D. $\cos^{-1}x$

Answer 1

Answer: (C)

$2\cos^{-1}(|x|)$

Answer 2

Let $\theta = \cos^{-1}x$. Then $x = \cos\theta$, where $0 \le \theta \le \pi$. Substituting $x = \cos\theta$ into the given equation: $y = \cos^{-1}(2(\cos\theta)^2 - 1)$ $y = \cos^{-1}(2\cos^2\theta - 1)$

Using the double angle identity $\cos(2\theta) = 2\cos^2\theta - 1$: $y = \cos^{-1}(\cos(2\theta))$

We know that $\cos^{-1}(\cos z) = z$ if $0 \le z \le \pi$, and $\cos^{-1}(\cos z) = 2\pi - z$ if $\pi < z \le 2\pi$. In our case, $z = 2\theta$. Since $0 \le \theta \le \pi$, the range of $2\theta$ is $[0, 2\pi]$.

Case 1: $0 \le 2\theta \le \pi$. This implies $0 \le \theta \le \frac{\pi}{2}$. Since $\theta = \cos^{-1}x$, this means $0 \le \cos^{-1}x \le \frac{\pi}{2}$, which corresponds to $0 \le x \le 1$. In this case, $y = 2\theta = 2\cos^{-1}x$.

Case 2: $\pi < 2\theta \le 2\pi$. This implies $\frac{\pi}{2} < \theta \le \pi$. Since $\theta = \cos^{-1}x$, this means $\frac{\pi}{2} < \cos^{-1}x \le \pi$, which corresponds to $-1 \le x < 0$. In this case, $y = 2\pi - 2\theta = 2\pi - 2\cos^{-1}x$.

Combining both cases, we can write the solution piecewise: $y = \begin{cases} 2\cos^{-1}x & \text{if } 0 \le x \le 1 \\ 2\pi - 2\cos^{-1}x & \text{if } -1 \le x < 0 \end{cases}$

We can express this more compactly using the absolute value function. For $x \in [-1, 1]$: If $x \ge 0$, then $|x|=x$, and $2\cos^{-1}(|x|) = 2\cos^{-1}x$. If $x < 0$, then $|x|=-x$, and $2\cos^{-1}(|x|) = 2\cos^{-1}(-x)$. Using the property $\cos^{-1}(-x) = \pi - \cos^{-1}x$, we get $2(\pi - \cos^{-1}x) = 2\pi - 2\cos^{-1}x$.

Thus, the expression $y = 2\cos^{-1}(|x|)$ covers both cases.