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

  • $\begingroup$ $$1- (2\cos^2x-1) = 1-2*\cos^2x+1= (2-2\cos^2x)$$ was this your question ? Not sure if i understood it Right. $\endgroup$ – user317721 May 14 '16 at 8:47
  • $\begingroup$ Hi @residuence, I've amended a bit my question. see the equivalence $\endgroup$ – Andy K May 14 '16 at 8:49
  • $\begingroup$ The identity is true, but you need to explain stepwise how you reached $2 - 2 cos^2 x$ for the $LHS$ $\endgroup$ – true blue anil May 14 '16 at 8:49
  • $\begingroup$ Hi @trueblueanil let me do that before I go to do the chores. What I know is how I can reach the equivalence... $\endgroup$ – Andy K May 14 '16 at 8:51
  • 2
    $\begingroup$ 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 $-2\cos^2x + 1$. $\endgroup$ – Arthur May 14 '16 at 8:51

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.