# Thread: Reversing the inequality direction, less than to greater and equal to

1. ## Reversing the inequality direction, less than to greater and equal to

Hi Forum
Using intervals and inequalities can get a little tricky, so here is my question.
Find all of the degree-measure of all the angles x in the interval $\displaystyle 0º\leq x \leq 360º$ for which
$\displaystyle 2(sin{x}+cos{x})<\sqrt{1+2sin{x}cos{x}}$
$\displaystyle 4(sin{2x}+1)<1+sin{2x}$
$\displaystyle sin{2x}<\frac{-3}{3}$
$\displaystyle sin{2x}+1\color{blue}{\geq} 0$

How could this happen?
I know that the sign of the inequality change when dealing with negative numbers or reciprocals, but...
This actually becomes greater then or equal. Can someone explain?
Thanks!

2. ## Re: Reversing the inequality direction, less than to greater and equal to

Originally Posted by Zellator
$\displaystyle 4(sin{2x}+1)<1+sin{2x}$
$\displaystyle sin{2x}<\frac{-3}{3}$
How could this happen?
First, that whole process makes no sense to me.
$\displaystyle sin{2x}<\frac{-3}{3}=-1$ is false for every x.

The sine function has no values less than -1.

3. ## Re: Reversing the inequality direction, less than to greater and equal to

Now that you mention it, Plato
The solution says
$\displaystyle 2(sinx+cosx)<\sqrt{1+{sinx}+{cosx}} \rightarrow sin2x+1\geq 0$
which is only true when $\displaystyle sinx+cosx<0$.
But
$\displaystyle sinx+cosx=\sqrt{2} sin{(x+45^\circ)}$
so
$\displaystyle 135^\circ<x<315^\circ$

It doesn't make sense this way, but reversing the direction we get something.
Have you check my steps? Is there something amiss?
Thanks Plato.

4. ## Re: Reversing the inequality direction, less than to greater and equal to

I'd prefer to simplify the RHS using algebra

$\displaystyle 1 + 2\sin(x)\cos(x) = (\sin(x)+\cos(x))^2$ which means the RHS becomes $\displaystyle \sqrt{1+2\sin(x)\cos(x)} = \sqrt{(\sin(x) + \cos(x))^2}$

$\displaystyle 2(\sin(x) + \cos(x)) < |\sin(x) + \cos(x)|$

$\displaystyle 2(\sin(x)+\cos(x)) - |(\sin(x) + \cos(x))| < 0$

Hence either $\displaystyle 3(\sin(x)+\cos(x)) < 0 \text{ or } \sin(x)+\cos(x) < 0$

sin and cos are both positive in the first quadrant so we can discard that as a solution

In quadrant 2 sin is positive but cos is negative so in this quadrant if $\displaystyle \cos(x) > \sin(x)$ then the inequality is satisfied, 135 degrees is this quadrant so $\displaystyle 135^o < x \leq 180^0$ works.

In quadrant 3 sin and cos are both negative so their sum will also be negative, hence all of quadrant 3 satisfies.

In quadrant 4 sin is negative and cos positive so the opposite of quadrant 2 applies and we want the first half of the quadrant - namely $\displaystyle 270 \leq x < 315$

Combining we get $\displaystyle 135^o < x < 315^o$.

No doubt there is a more efficient way though

5. ## Re: Reversing the inequality direction, less than to greater and equal to

Originally Posted by e^(i*pi)
I'd prefer to simplify the RHS using algebra

$\displaystyle 1 + 2\sin(x)\cos(x) = (\sin(x)+\cos(x))^2$

which means the RHS becomes $\displaystyle \sqrt{1+2\sin(x)\cos(x)} = \sqrt{(\sin(x) + \cos(x))^2}$
$\displaystyle 135^o < x < 315^o$.

Hi e^(i*pi)!
There may be some othe way to solve this but, this is great, no doubt about that.

Just one thing bothers me
When you squared the equation you didn't forget the 2 in $\displaystyle {\color{blue}2^2}(\sin(x)+\cos(x))^2$ ?

Later you don't use the 4 anyway, right? (I need confirmation on this one)
Since you only considered $\displaystyle (\sin(x)+\cos(x))<0$

I'll keep this solution in mind, thanks for your help, so far!

6. ## Re: Reversing the inequality direction, less than to greater and equal to

Originally Posted by Zellator
Hi e^(i*pi)!
Just one thing bothers me
When you squared the equation you didn't forget the 2 in $\displaystyle {\color{blue}2^2}(\sin(x)+\cos(x))^2$ ?
I didn't square anything. I rewrote the RHS and left the LHS.

I used algebra to show that $\displaystyle \sqrt{1+2\sin(x)\cos(x)} = \sqrt{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|$. This is because $\displaystyle \sqrt{a^2} = |a|$

This is equal to the unchanged LHS: $\displaystyle 2(\sin(x)+\cos(x)) = |\sin(x) + \cos(x)|$. The square root cancelled of it's own accord due to the algebraic manipulation of the RHS.

Later you don't use the 4 anyway, right? (I need confirmation on this one)
Since you only considered $\displaystyle (\sin(x)+\cos(x))<0$

I'll keep this solution in mind, thanks for your help, so far!
Yes, there are two equations that would need solving but since they only differ by amplitude their intervals would be equal to each other

7. ## Re: Reversing the inequality direction, less than to greater and equal to

Originally Posted by e^(i*pi)
I didn't square anything. I rewrote the RHS and left the LHS.

I used algebra to show that $\displaystyle \sqrt{1+2\sin(x)\cos(x)} = \sqrt{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|$. This is because $\displaystyle \sqrt{a^2} = |a|$

This is equal to the unchanged LHS: $\displaystyle 2(\sin(x)+\cos(x)) = |\sin(x) + \cos(x)|$. The square root cancelled of it's own accord due to the algebraic manipulation of the RHS.

Yes, there are two equations that would need solving but since they only differ by amplitude their intervals would be equal to each other
I see now, thanks!
It's a great idea to use 1 as cos^2x+sen^2x and use it for factoring.
Thanks for explaining why you disconsidered one of the results, that's a great help indeed.
I'll put the information to good use!
Thanks e^(i*pi)!

All the best!