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 for which
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!
Now that you mention it, Plato
The solution says
which is only true when .
But
so
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.
Your help is really appreciated!
I'd prefer to simplify the RHS using algebra
which means the RHS becomes
Hence either
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 then the inequality is satisfied, 135 degrees is this quadrant so 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
Combining we get .
No doubt there is a more efficient way though
Hi e^(i*pi)!
Thanks so much for your reply!
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 ?
Later you don't use the 4 anyway, right? (I need confirmation on this one)
Since you only considered
I'll keep this solution in mind, thanks for your help, so far!
I didn't square anything. I rewrote the RHS and left the LHS.
I used algebra to show that . This is because
This is equal to the unchanged LHS: . 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 otherLater you don't use the 4 anyway, right? (I need confirmation on this one)
Since you only considered
I'll keep this solution in mind, thanks for your help, so far!