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Can we write $\sin x > a$ as $x > \arcsin a$. Please explain the process. Is it possible for all ratios with any inequality sign.

user1729
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3 Answers3

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Since $\sin x$ is as well as $\arcsin x$ increasing, we can write that* $x>\arcsin a$ if $\sin x>a$.

*Note that this is not always true, consider the sequence of equations: $$\sin x>a\\\arcsin(\sin x)>\arcsin(a)\\x>\arcsin a\text{ if }x\in[-\pi/2,\pi/2]\wedge a\in[-1,1]$$ because a must satisfy $\arcsin$'s domain demand and $\arcsin\sin x$ is not always $x$. This is the graph of $\arcsin\sin x$ which is exactly equal to x in $[-\pi/2,\pi/2]$. enter image description here

So you'll end up for $x\in[\pi/2,3\pi/2]$ $$\arcsin(\sin x)>\arcsin(a)\\ \pi-x>\arcsin(a)$$ Not the thing you mentioned.Other similiar functions are $\tan$ and $\sec$.

Also, if we worked with cos, you'll end up something like this: $$\cos x>a\\\arccos(\cos x)<\arccos(a)\\x<\arccos a\text{ if }x\in[-\pi/2,\pi/2]\wedge a\in[-1,1]$$ The reason I have inverted the inequality is that $cos x$ is increasing whereas $\arccos x$ is decreasing.Other similiar cases are $\cot$ and $\csc x$.

RE60K
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  • @user168467 if this or any other answer solved your problem or is useful you can accept it or upvote it, if you didn't knew. – RE60K Aug 07 '14 at 14:27
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Strictly Decreasing Function

If $f(\beta)<f(\alpha)$ for all $\beta > \alpha$, then $f(x)$ is a strictly decreasing function.

All you have to do is determine if the function in question is strictly decreasing or not. If it's strictly decreasing then the direction of the inequality changes. Otherwise, the direction remains the same.

Examples

\[ f(x)=\arcsin(x) \] $f(x)$ is not a strictly decreasing function, the direction of the inequality stays the same.

\[ f(x)=\arccos(x) \] $f(x)$ is a strictly decreasing function, so the direction of the inequality changes.

\[ f(x)=-x \] $f(x)$ is a strictly decreasing function. This is why the direction of an inequality changes when you multiply or divide an inequality by a negative number.

k170
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Only for values of x in the first and fourth quadrants -- which is to say the fundamental domain of arcsin. Something similar works for all trigonometric functions in the first quadrant; sin tan and sec keep the inequality direction the same; cos cot and csc switch it.

Dan Uznanski
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