Can You Square an Inequality for Sin(x)?

[kahwawashay1]

Hello

I am doing a calculus proof with epsilon-delta and I am trying to say the following:

-1\leqsin x\leq1

and now I want to get (sin x )^2 ...so can you just square all sides of the inequality like this:

(-1)^2\leq(sin x)^2\leq(1)^2

??

According to the rule for inequalities, you can do this i think? But obviously sinx squared isn't between 1 and 1?

 

[Tomer]

Hello

I am doing a calculus proof with epsilon-delta and I am trying to say the following:

-1\leqsin x\leq1

and now I want to get (sin x )^2 ...so can you just square all sides of the inequality like this:

(-1)^2\leq(sin x)^2\leq(1)^2

??

According to the rule for inequalities, you can do this i think? But obviously sinx squared isn't between 1 and 1?

Well, you obviously can't :-)

You can only square an inequality if you know that all the expressions in it are positive.
In this case -1 isn't positive, so...

 

[micromass]

No, you can't do this. The rule

a\leq b~\Rightarrow~a^2\leq b^2

only holds if a,b\geq 0.

If both a,b\leq 0, then we got the reverse rule

a\leq b~\Rightarrow b^2\leq a^2

If we have a\leq 0\leq b then all sort of things can happen. It's not possible to find a relation between a^2 and b^2 just like that.

 

[Char. Limit]

Another possible thing to do would be to multiply by sin(x), which is totally viable for x \in \left[0, \pi\right].

 

[LCKurtz]

Would squaring the inequality
0\leq |sin(x)|\leq 1
help you?

 

[kahwawashay1]

Would squaring the inequality
0\leq |sin(x)|\leq 1
help you?

Yup this helps thx!