# Thread: Whoah! what is going on with this identity!

1. ## Whoah! what is going on with this identity!

(1-sin2x) = sinx-cosx
sinx-cosx

2. Hello,

$\sin(2x)=2\sin(x)\cos(x)=\sin(x)\cos(x)+\sin(x)\co s(x)$

$1=\cos^2(x)+\sin^2(x)$

Hence

$1-\sin(2x)=\cos^2(x)+\sin^2(x)-\sin(x)\cos(x)-\sin(x)\cos(x)$

$=\cos^2(x)-\sin(x)\cos(x)+\sin^2(x)-\sin(x)\cos(x)$

$=\cos(x)(\cos(x)-\sin(x))+\sin(x)(\sin(x)-\cos(x))$

$=-\cos(x)(\sin(x)-\cos(x))+\sin(x)(\sin(x)-\cos(x))$

$=(\sin(x)-\cos(x))(\sin(x)-\cos(x))$

So what else ?

3. Originally Posted by Jono Mathematician
(1-sin2x) = sinx-cosx
sinx-cosx
Hint: Note that $1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2$ ....oh well, I just killed the problem....

EDIT: plus, Moo beat me to it!

4. Oh well, why did I look for the difficult way ?

5. please can u write it in like fraction form as i have no clue as to how to read it haha..

ps thanks

6. nevermind, haha just got it ... please can you try my other one ... as i dunno what to do there ... it is also by Jono Mathematician thanks

7. Originally Posted by Jono Mathematician
please can u write it in like fraction form as i have no clue as to how to read it haha..

ps thanks
hehe, nice try. But we did not do the problem for you, we gave you a hint. The important part is to realize that $1 - \sin 2x = (\sin x - \cos x)^2$

Now you try the problem

8. 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

9. haha, once again i just figured it out, thanks

10. Originally Posted by Jono Mathematician
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 $1 - \sin 2x$, write what i told you it is equal to