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?
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....
= cosx sin2x/ sinx
= [cosx (2sinx cosx)] / sinx
And then I'm lost...
Where'd the extra 2 come from? It disappears later, but two wrongs don't make a right.On the right hand side I have:
= cosx/sinx (sin2x)
= (cosx/sinx) (2sinx cosx / 1)
= 2sinx 2cos^2x / sin x
You didn't cancel the sin(x).= sinx 2cos^2x
What does your proof look like now?But that's clearly not right....
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
Pretty much.and then 2sinx/sinx would just leave the 2 somehow to get 2cos^2x??
It would equal 2, actually.I'm a little confused. I thought 2sinx/sinx would equal sinx
You're almost there! I would just clean up your notation a little and present it better.