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Math Help - Whoah! what is going on with this identity!

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

    (1-sin2x) = sinx-cosx
    sinx-cosx
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  2. #2
    Moo
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    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 ?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    (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!
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  4. #4
    Moo
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    Oh well, why did I look for the difficult way ?
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  5. #5
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    please can u write it in like fraction form as i have no clue as to how to read it haha..

    ps thanks
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  6. #6
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    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
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    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
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  8. #8
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    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
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  9. #9
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    haha, once again i just figured it out, thanks
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    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
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