Is it possible that $2-2\cos^2x$ is equivalent to $1-(2\cos^2x-1)$

There is this exercise and for the first time in my life, I don't want to go to see the solution. Instead, I'm more asking of a tiny help to see if I'm right in my conclusion

Kids are getting concerned about this math fascination and I said to them it is for their good...

Humor aside, let's crack on. This is my equation below:

$$\sin 2x \tan x = 1 - \cos( 2x)$$

$RHS$ is equal to $1- (2\cos^2x-1)$

So I started $LHS$ and eventually, I found this below

$$(2-2\cos^2x)$$

I'm not good in maths as many of you but intuitively, I can yet feel that my solution is equivalent to the $RHS$ I mentioned above. However, something is missing to me but I cannot say where...

$$(2-2\cos^2x) \equiv 1- (2\cos^2x-1)$$

Thanks again for your patience in bearing with me

• $$1- (2\cos^2x-1) = 1-2*\cos^2x+1= (2-2\cos^2x)$$ was this your question ? Not sure if i understood it Right. – user317721 May 14 '16 at 8:47
• Hi @residuence, I've amended a bit my question. see the equivalence – Andy K May 14 '16 at 8:49
• The identity is true, but you need to explain stepwise how you reached $2 - 2 cos^2 x$ for the $LHS$ – true blue anil May 14 '16 at 8:49
• Hi @trueblueanil let me do that before I go to do the chores. What I know is how I can reach the equivalence... – Andy K May 14 '16 at 8:51
• The two terms are equal (equivalent), but the reason why has nothing to do with trigonometry and all to do with basic algebra. Note that $-(2\cos^2 x - 1)$ is exactly the same as $-2\cos^2x + 1$. – Arthur May 14 '16 at 8:51

$$1- (2\cos^2x-1) = 1-2*\cos^2x+1= (2-2\cos^2x)$$ Greetings.