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Math Help - Trig Functions...Simplifying..urgent help needed

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    Trig Functions...Simplifying..urgent help needed

    How do I simplify these trig functions so that the right and left sides are the same?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    How do I simplify these trig functions so that the right and left sides are the same?
    you could note that both sides are \cos 2x, but that would be too easy.

    instead, use the fact that \sin^2 x + \cos^2 x = 1. start with the left hand side of the equation and use the identity i typed abpve to replace either the sine or cosine to get the right hand side
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    Do I do this for both problems?
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    for number 5, I get
    cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    for number 5, I get
    cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x
    no.

    start with the left hand side only. forget about the right. then use the identity i showed you to transform it into the right hand side (or vice versa, start with the right side and simplify to get the left)
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    I did it this way, I don't know if this is correct, but it made sense and worked out...is this what you meant?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    I did it this way, I don't know if this is correct, but it made sense and worked out...is this what you meant?
    no, it's not correct. you still did not do what i asked.

    \cos^2 x - \sin^2 x = 1 - 2\sin^2 x

    consider the LHS

    \cos^2 x - \sin^2 x = \left( 1 - \sin^2 x \right) - \sin^2 x ...........by the identity i showed you

    = 1 - 2 \sin^2 x

    = RHS

    or, we could consider the RHS:

    1 - 2 \sin^2 x = \sin^2 x + \cos^2 x - 2 \sin^2 x

    = \cos^2 x = \sin^2 x

    = LHS

    either way, we see that \cos^2 x - \sin^2 x = 1 = 2 \sin^2 x

    but you could just as easily simplify both sides to \cos 2x
    Last edited by Jhevon; November 26th 2007 at 07:08 PM. Reason: Fixed LaTex error
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    I think that I have got it....but on the second problem, I am missing -1 in my final solution....
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    Quote Originally Posted by aikenfan View Post
    I think that I have got it....but on the second problem, I am missing -1 in my final solution....
    what you did is incorrect. 1 - 2sin^2(x) is not 2cos^2(x)

    here you should have changed the sin^2(x) into 1 - cos^2(x). you are aiming for something with only cos^2(x), why change the cosine to sine?
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    So do I take the 1 - sin^x - sin^2x and change it to:
    1 - sin^2x - 1 - cos^2x?
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    Quote Originally Posted by aikenfan View Post
    So do I take the 1 - sin^x - sin^2x and change it to:
    1 - sin^2x - 1 - cos^2x?
    no, change the sin^2 x into 1 - cos^2 x
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    Ok, I think that I have got it...i now have 1 - cos^2x - cos^x
    and that simplifies to 1 - 2cos^2x ?
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    Quote Originally Posted by aikenfan View Post
    Ok, I think that I have got it...i now have 1 - cos^2x - cos^x
    and that simplifies to 1 - 2cos^2x ?
    \cos^2 x - \sin^2 x = \cos^2 x - \left( 1 - \cos^2 x \right)

    = \cos^2 x - 1 + \cos^2 x

    = 2 \cos^2 x - 1

    as desired
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