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Math Help - [SOLVED] and what is going on with this one as well!?!

  1. #1
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    [SOLVED] and what is going on with this one as well!?!

    cos2x
    (cosx+sinx)^3

    =

    cosx-sinx
    1+sin2x
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    cos2x
    (cosx+sinx)^3

    =

    cosx-sinx
    1+sin2x
    Start with the top function. the trick here is to notice that \cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)

    Now try again
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  3. #3
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    Thank you so much, it is sometimes confusing as to what to use in place of cos2x as there are three options:

    cos^2x - sin^2x
    2cos^2x - 1
    1-2sin^2x

    so it is difficult to choose which one is appropriate. Is there any hint as to which on to use during what type of equation, as in which cos2x double angle should you use when dealing with different sums?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    Thank you so much, it is sometimes confusing as to what to use in place of cos2x as there are three options:

    cos^2x - sin^2x
    2cos^2x - 1
    1-2sin^2x

    so it is difficult to choose which one is appropriate. Is there any hint as to which on to use during what type of equation, as in which cos2x double angle should you use when dealing with different sums?
    you have to look at what you are given. i used the first form because of the \cos x - \sin x that was in the other function. of course this reminds you of \cos^2 x - \sin^2 x, but the squares were a problem. how do we get rid of them. well, we could realize that it is the difference of two squares and hence can be written as (\cos x - \sin x)(\cos x + \sin x), and hence we get a factor in the other function, which is good
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