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Assuming this is correct:

$\arcsin x \neq 0$

$\sin(0) \neq x$

$0\neq x$

Following the same logic, why is this incorrect?

$\arcsin(x+\frac{1}{3}) \geq 0$

$\sin (0) \geq x+\frac{1}{3}$

$0 \geq x + \frac{1}{3}$

$x \leq -\frac{1}{3}$

$x\in(-\infty; -\frac{1}{3}] $

The correct answer should be:

$x\in[-\frac{1}{3}; \infty)$

Why is this incorrect? Can I use the $ \sin(0) $ trick only for when there's $\neq$ sign, thus cannot be used in inequalities?

I do have one more question, by the way. Is this correct?

$ arcsin(expression) \geq 0 $

$ expression\geq 0 $

and

$arccos (expression) \geq 0 $

$ expression\geq 0 $

Thanks

weno
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1 Answers1

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The arcsine is an increasing function, so $\arcsin a\ge\arcsin b$ is equivalent to $a\ge b$ (provided $a$ and $b$ are in the domain of the arcsine).

Thus your inequality $\arcsin(x+1/3)\ge0=\arcsin 0$ becomes $x+1/3\ge0$, that is, $x\ge-1/3$. Taking into account that $-1\le x+1/3\le 1$, we get $-4/3\le x\le 2/3$, so combining the inequalities yields $$ -\frac{1}{3}\le x\le\frac{2}{3} $$ There's no way the solution set is $[-1/3;\infty)$

For the arccosine you have to take into account that it is decreasing, so $\arccos a\ge\arccos b$ becomes $a\le b$. Since $0=\arccos(1)$, the inequality $\arccos x\ge 0$ becomes $x\le1$. This is anyway obvious, because $\arccos x\ge0$ for every $x\in[-1;1]$.

egreg
  • 238,574