# Trigometric Identies Rearrangement

1. Jun 3, 2008

### thomas49th

1. The problem statement, all variables and given/known data

I have the term:

2tan2x + (2x-1)(2sec²2x) = 0

I need to rearrange it into the form

4x + sin4x - 2 = 0

I tried:
$$\frac{2sin2x}{cos2x} + \frac{2(2x-1)}{cos²2x} = 0$$

i multiply each side by cos²2x

$$2.Sin2x.Cos2x + 4x - 2= 0$$

but now where?

Thanks

2. Jun 3, 2008

### Vid

Look at the sine double angle identity.

3. Jun 3, 2008

### thomas49th

i thought about that but you cant use it... can you?

sin2x = 2sinxcosx

you could do

2(2sinxcosx)cos2x + 4x - 2

or can you do somthing with the 2sin2xCos2x ??? it's 2x not x so you cant equate it to sin2x

Not sure :/

Thanks

4. Jun 3, 2008

### Vid

Let u = 2x

2sin(u)cos(u) = sin(2u)

5. Jun 3, 2008

### thomas49th

but to equal sin4x doesnt it need to be 4sinxcosx not 2sin2xcos2x?

Thanks

6. Jun 3, 2008

### Vid

Not at all.
Obviously 2sin(u)cos(u) = sin(2u) is true since the variable inside the formula doesn't matter. Now, replace u with 2x.
2sin(2x)cos(2x) = sin(2*(2x)) = sin(4x).
There's no need for another 2 out in front.

7. Jun 3, 2008

### thomas49th

and that's ture for all double angle formulas
and similar for the half angle formulas?

Cheerz :)