(1-sin2x) = sinx-cosx
sinx-cosx
Hello,
$\displaystyle \sin(2x)=2\sin(x)\cos(x)=\sin(x)\cos(x)+\sin(x)\co s(x)$
$\displaystyle 1=\cos^2(x)+\sin^2(x)$
Hence
$\displaystyle 1-\sin(2x)=\cos^2(x)+\sin^2(x)-\sin(x)\cos(x)-\sin(x)\cos(x)$
$\displaystyle =\cos^2(x)-\sin(x)\cos(x)+\sin^2(x)-\sin(x)\cos(x)$
$\displaystyle =\cos(x)(\cos(x)-\sin(x))+\sin(x)(\sin(x)-\cos(x))$
$\displaystyle =-\cos(x)(\sin(x)-\cos(x))+\sin(x)(\sin(x)-\cos(x))$
$\displaystyle =(\sin(x)-\cos(x))(\sin(x)-\cos(x))$
So what else ?
i got it but cannot understand how you got rid of the -cosx+sinx
the answer doesnt equal (sinx-cosx)^2... from that last step, how did you get from -cosx(sinx-cosx) etc to:
(sinx-cosx)+(sinx-cosx)? i mean where did the -cosx before the bracket go to? as well as the +sinx before the bracket?
thanks
i honestly do not know what you are talking about. I looked back at my post and realized that i did nothing that you said i did. apparently you are confused about something and are reading what i did incorrectly. for starters, rewrite the question, but instead of writing $\displaystyle 1 - \sin 2x$, write what i told you it is equal to