trig identity proof - need help

[msfior]

Hi, I have to proove:

tan2x-sin2x = (sin2xtan2x)/cotx

I start with the right side as it is more complicated and cot can be expressed as cosx/sinx. But I'm lost on trying to turn this into the left side...

Any help would be appreciated!

Thanks

[g***@nukove.com]

Try starting with the left side, using the fact that

tan2x = sin2x/cos2x

[msfior]

Where do I go from there once I have:

tan2x-sin2x=sin2x/cos2x - sin2x

Do I make a common denom or do I do the double angle formulas???

[mathman]

LHS: tan2x-sin2x=tan2x(1-cos2x)

Therefore we need to show:
1-cos2x=sin2x/cotx (or sin2xtanx)

LHS: 1-cos2x=1-cos(^2)x+sin(^2)=2sin(^2)x

RHS: sin2x tanx = 2sinxcosx sinx/cosx = 2sin(^2)x

[msfior]

How did you get this:
LHS: tan2x-sin2x=tan2x(1-cos2x)

[Michael Varney]

Post by msfior
tan2x-sin2x = (sin2xtan2x)/cotx
I start with the right side as it is more complicated and cot can be expressed as cosx/sinx. But I'm lost on trying to turn this into the left side...
Any help would be appreciated!
Thanks

Try it with exp.

[msfior]

can you elaborate on that ??? what do you mean?

[Michael Varney]

Post by msfior
can you elaborate on that ??? what do you mean?

Exp[i x] = Cos[x] + i Sin[x]

From that relation you can prove nearly all the trig identities.

[msfior]

I see... I haven't learned that and I think it's beyond the scope of my current course!

Thanks...

[Anon]

Post by msfior
tan2x-sin2x = (sin2xtan2x)/cotx
I start with the right side as it is more complicated and cot can be
expressed as cosx/sinx. But I'm lost on trying to turn this into the
left side...
Any help would be appreciated!
Thanks
LHS: tan2x-sin2x=tan2x(1-cos2x)
1-cos2x=sin2x/cotx (or sin2xtanx)
LHS: 1-cos2x=1-cos(^2)x+sin(^2)=2sin(^2)x
RHS: sin2x tanx = 2sinxcosx sinx/cosx = 2sin(^2)x

The previous post from 'mathman' provided one proof which apparently
you did not follow. I'll just add a bit of detail to that.

LHS: (tan 2x) - (sin 2x)
RHS: (sin 2x) (tan 2x) / (cot x)
--------------------------------

You know that tan 2x = (sin 2x) / (cos 2x)
and so sin 2x = (tan 2x) (cos 2x)

Substitute for (sin 2x) in the LHS, giving

(tan 2x) - (tan 2x) (cos 2x) = (tan 2x) [1 - cos 2x]


Now, equate this and the RHS, cancel the common factor (tan 2x), and
you get

1 - cos 2x = (sin 2x) / (cot x) (*)

You know that
cos 2x = 1 - 2 (sin x)^2
sin 2x = 2 (sin x) (cos x)
1 / (cot x) = tan x = (sin x) / (cos x)

Substitute these in (*) above and you get

1 - [1 - 2 (sin x)^2] = [2 (sin x) (cos x)] [(sin x) / (cos x)]
2 (sin x)^2 = 2 (sin x)^2

Done, HTH.

[Z***@pidham.vispa.com]

Try using the tan formulae for double angles:

if t = tan(x)

tan(2x) = 2t/(1 - t^2)
sin(2x) = 2t/(1 + t^2)
cos(2x) = (1 - t^2)/(1 + t^2)

It drops out in a couple of lines.

Jon