# Squaring sine in inequality?

1. Sep 26, 2011

### kahwawashay1

Hello

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

-1$\leq$sin 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 isnt between 1 and 1?

2. Sep 26, 2011

### Tomer

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...

3. Sep 26, 2011

### micromass

Staff Emeritus
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.

4. Sep 26, 2011

### Char. Limit

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

5. Sep 26, 2011

### LCKurtz

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

6. Sep 26, 2011

### kahwawashay1

Yup this helps thx!

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