1. ## Proving Another Identity

I'm stuck again... I don't know why I am having so much trouble with these. There is clearly something I'm just not seeing.

1 + cos2x = cotx sin2x

I think I need to simplify the right hand side?

This is what I've done, but I'm making a real mess of things....

cotx sin2x
= (cosx/sinx)(sin2x)
= cosx sin2x/ sinx
= [cosx (2sinx cosx)] / sinx

And then I'm lost...

2. In problems such as this, the first thing I would do is get all trig functions to have the same argument. I never know what I have until I do that. Use the double-angle formulas to get rid of the 2x arguments. What do you get?

3. On the left hand side I have:

1 + cos2x
= 1 + 2cos^2 - 1
= 2cos^2x

On the right hand side I have:

cotx sin2x
= cosx/sinx (sin2x)
= (cosx/sinx) (2sinx cosx / 1)
= 2sinx 2cos^2x / sin x
= sinx 2cos^2x

But that's clearly not right....

4. Originally Posted by starshine84
On the left hand side I have:

1 + cos2x
= 1 + 2cos^2 - 1
= 2cos^2x
Looking good so far.

On the right hand side I have:

cotx sin2x
= cosx/sinx (sin2x)
= (cosx/sinx) (2sinx cosx / 1)
= 2sinx 2cos^2x / sin x
Where'd the extra 2 come from? It disappears later, but two wrongs don't make a right.

= sinx 2cos^2x
You didn't cancel the sin(x).

But that's clearly not right....
What does your proof look like now?

5. I don't think I'm following you on the right hand side...

If I'm multiplying the numerators, does cosx (2sinx cosx) = 2sinx cos^2. and then 2sinx/sinx would just leave the 2 somehow to get 2cos^2x?? I'm a little confused. I thought 2sinx/sinx would equal sinx

6. Originally Posted by starshine84
I don't think I'm following you on the right hand side...

If I'm multiplying the numerators, does cosx (2sinx cosx) = 2sinx cos^2.
Well, cos(x) ( 2 sin(x) cos(x) ) = 2 sin(x) cos^2(x) is a bit clearer way to write it (I'm a stickler for parentheses around all function arguments), but I think you've got the idea there.

and then 2sinx/sinx would just leave the 2 somehow to get 2cos^2x??
Pretty much.

I'm a little confused. I thought 2sinx/sinx would equal sinx
It would equal 2, actually.

You're almost there! I would just clean up your notation a little and present it better.

7. Okay, I think I'm with you. Thank you again!! This unit is giving me a lot of trouble for some reason.

8. You're welcome!