[sally aa]

proof that left side = right side

[Unknown008]

First thing to notice is that the left side has double angles whereas the right hasn't. So, you can start by breaking the right side down to single angles.

[sally aa]

Well I try and I couldn't solve it
I try both sides and I couldn't get the right answer

[Unknown008]

Could you post what you tried?

[sally aa]

(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-2(sinx)^2)

[sally aa]

I try this side and on the half way I stopped

[Unknown008]

(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-(sinx)^2)

Remember that:

[sally aa]

2(sinx)^2 =2sin^2(x) yes I know

[Unknown008]

Okay, now expand the denominator. What do you get?

[sally aa]

(2sinxcosx)(2sinxcosx)/1-1+2sin^2(x)
=(2sinxcosx)(2sinxcosx)/2sin^2(x)

[Unknown008]

Good. Now what can you cross out?

[sally aa]

(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?

[sally aa]

(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) that's what I get

[Unknown008]

(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?

I don't understand what you mean here...

[Unknown008]

(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) thats what I get

The bolded part is wrong.

[sally aa]

What do u mean ?

[Unknown008]

I mean:

[sally aa]

sorry its 4sin^2(x)cos^2(x)

[Unknown008]

Okay, so now you have:



What can you simplify?

[sally aa]

4sin^2(x)cos^2(x)/2sin^2(x) then if I canceled the 4sin^2(x) with 2 sin^2(x) we end up with 2 cos^2(x) is this right ?