Can someone show all steps necessary to verify the trigonometric identity? I don't really know how to do it on my own. Thanks! 1 - 2cos^2x = 2sin^2x - 1
\[1-2\cos^2x = 2\sin^2x -1\]
In case it didn't make sense the first time?
use \(\huge \color{red}{\sin^2 x+\cos^2x=1}\)
I honestly have no idea how....
ok. from that \(\huge \color{red}{\sin^2 x=1-\cos^2x}\) so substitute this for sin^2 x on right side. what u get ? 2(...?...)-1
got that ^ ?
Do you get cos^2x?
lets see what we get right side = 2sin^2 x -1 =\(\huge \color{red}{2(1-\cos^2x)-1}\) =..... ? can u tell me next step ?
so it would be 2-2cos^2x -1 which would end up being 1-2cos^2x Right..? & Thank you!
absolutely correct :)
and \[ \begin{array}l\color{red}{\text{w}}\color{orange}{\text{e}}\color{#e6e600}{\text{l}}\color{green}{\text{c}}\color{blue}{\text{o}}\color{purple}{\text{m}}\color{purple}{\text{e}}\color{red}{\text{ }}\color{orange}{\text{^}}\color{#e6e600}{\text{_}}\color{green}{\text{^}}\color{blue}{\text{}}\end{array} \]
& That was on our right side, so now they equal each other! Yay! :) How did you know to use the sin^2x + cos^2x=1? I know that not squared they equal 1, but even when they are squared they equal 1? Does my question make sense?
I meant I knew without the 2 in front of them they equaled 1.
Like that I knew it equaled 1
no, your Q didn't make sense to me, sorry.
that is squared..
Hmm. let me try again. So no matter the coefficent attatched to the front of sin^2theta _ cos^2theta, the answer is one?
I meant coefficient not squared... Whoops!
Dang it! I meant + not _
no, the co=efficients would matter. if they are different , it will not be =1 but if the co-efficient are same, we can factor out that co-efficient, then it =1 like example asin^2x+bcos^2x is not 1 but asin^2x+acos^2x = a(1)
Ohh okay, I get it now. Thank you! & sorry for confusing you!
no problem :)
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