Functions #2

Geometry Level 3

$\large f(x) = \sqrt{\sin^{-1}|\sin x| - \cos^{-1}|\cos x|}$

Find the range of the function above.

The answer is 0.

1 solution

Chew-Seong Cheong
Dec 7, 2017

Since $|\sin x| \ge 0$ for all $x$ , The solution of $\sin^{-1} |\sin x| = x_1$ , where $x_1$ is the corresponding value in the first quadrant $\left[0, \frac \pi 2\right]$ . Similarly, $\cos^{-1} |\cos x| = x_1$ , where $x_1 \in \left[0, \frac \pi 2\right]$ . Therefore, $f(x) = \sqrt{\sin^{-1} |\sin x| - \cos^{-1} |\cos x|}$ $= \sqrt {x_1-x_1} = \boxed{0}$ for all $x$ .

$x$ does not necessarily $\in [0, \frac{π}{2}]$

A Former Brilliant Member - 3 years, 6 months ago

For instance, if $x=\dfrac{3π}{4}$ then $\sin^{-1} |\sin \frac{3π}{4}| \neq \frac{3π}{4}$

A Former Brilliant Member - 3 years, 6 months ago

$\sin^{-1}x$ is defined for $-\frac \pi 2 x \le \frac \pi 2$ .

Chew-Seong Cheong - 3 years, 6 months ago

@Chew-Seong Cheong – $\sin^{-1} x$ is defined for $-1 \le x \le 1$

A Former Brilliant Member - 3 years, 6 months ago

I have changed the solution.

Chew-Seong Cheong - 3 years, 6 months ago

Functions #2

The answer is 0.

1 solution

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