Prove the following identies? Cos2X

[mathfun]

where did i go wrong?

 

[Gene]

Mathisfun:
Where did cot(x) in the denominator go
=(cosx/sinx) - 2sinxcosx
Where did "/sin(x)" go
Where did sin²(x) come from
You can't multiply by sin(x) only in RHS
=cosx - 2sin²(x)cos(x)
Where did cos(x) go

= 1 - 2sin²(x)
Some may be just typos, but I can't follow it. Try putting something you know (x=30 maybe, or even x=0) and see what comes out for each of the lines you have typed. They should all give the same answer if your steps are correct.

 

[mathfun]

R.H.S = cotx - sin2x we can all agree with this

=(cosx/sinx) - 2sinxcosx trig identities: cotx = cosx/sinx right? another trig identity: sin2x = 2sinxcosx

now can we agree with that so far?

 

[Gene]

Mathfun: Pingu is having enough trouble by himself. Re-post YOUR first post as a new topic and we'll talk about it there, if you want to.

 

[mathfun]

Pingu: I'm sorry if i'm confusing u..but just looking at the other way is way too complicated. I don't understand why genee can't follow my way because it is a lot simpler if you have ur trig identity sheet infront of you instead of dividing the whole thing by cotx =.= however, i was just trying to help. nevertheless, i am not forcing you to look/follow my way and confuse you more.

 

[Pingu]

Can someone please give me the full answer with all the steps please.
Otherwise I won't be able to finish my homework on the Christmas break.
I have five more of these questions and I have been stuck on this question for days.
I have more math after that and I also have chemistry.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\cos (2x) &=& \frac{{\cot (x) - \sin (2x)}}{{\cot (x)}} \\
&=& \frac{{\frac{{\cos (x)}}{{\sin (x)}} - 2\sin (x)\cos (x)}}{{\frac{{\cos (x)}}{{\sin (x)}}}} \\
&=& \frac{{\cos (x) - 2\sin ^2 (x)\cos (x)}}{{\cos (x)}} \\
&=& 1 - 2\sin ^2 (x) \\
&=& \cos (2x) \\
\end{array}\)

 

[Pingu]

Thank you for helping me, I understand it now.