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    I have the following identity to prove.

    From LHS I did
    = 1 - cos2x
    = 1 - 1/cos2x
    = cos2x - 1 / cos2x

    How further can I go to reach the RHS?
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    (Original post by nwmyname;[url="tel:66445502")
    66445502[/url]]I have the following identity to prove.

    From LHS I did
    = 1 - cos2x
    = 1 - 1/cos2x
    = cos2x - 1 / cos2x

    How further can I go to reach the RHS?

    Don't do it that way. LHS equals sine squared. The RHS wants some squared in terms of cos and sine so the obvious thing here is to introduce tan. Tan squared times cos squared is sin squared so you can work from there.
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    (Original post by nwmyname;[url="tel:66445502")
    66445502[/url]]I have the following identity to prove.

    From LHS I did
    = 1 - cos2x
    = 1 - 1/cos2x
    = cos2x - 1 / cos2x

    How further can I go to reach the RHS?
    Also your working out is wrong. The second line does not equal the one above it
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    (Original post by RDKGames)
    Don't do it that way. LHS equals sine squared. The RHS wants some squared in terms of cos and sine so the obvious thing here is to introduce tan. Tan squared times cos squared is sin squared so you can work from there.
    oh. so in this case, work from RHS to LHS then instead?
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    (Original post by nwmyname)
    oh. so in this case, work from RHS to LHS then instead?
    No. RHS should just hint you to what you should getting towards from LHS.


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    (Original post by nwmyname)
    I have the following identity to prove.

    From LHS I did
    = 1 - cos2x
    = 1 - 1/cos2x
    = cos2x - 1 / cos2x

    How further can I go to reach the RHS?
    It is a standard result (which you should know) that sec^2x - 1 = tan^2x.
    Thus the RHS becomes tan^2x(1-sin^2x) = tan^2x(cos^2x) using the identity sin^2x+cos^2x=1
    and using the definition of tan as sin/cos, this becomes sin^2x.
    Also the LHS is just 1-cos^2x = sin^2x, so since both sides are equal to sin^2x, they must be equal as required. Simple.

    Note that this technique of working with both sides and "meeting in the middle" is very versatile and recommended to use when both sides of the identity to be proved are not in their simplest forms.
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    (Original post by nwmyname)
    Here is what I have done so far.

    1 - cos2x
    = 1 - 1/sec2x
    = 1 - 1/1+tan2x

    = (1 + tan2x - 1) / 1+tan2x
    = tan2x / 1 + tan2x

    = tan2x * 1/(1+tan2x)
    = (sec2x - 1) * 1/(1+tan2x)

    How further can I go?
    Overly long winded, but okay. Now notice that:

    \displaystyle \frac{1}{1 + \tan^2 x} = \frac{1}{1 + \frac{\sin^2 x}{\cos^2 x}} = \frac{\cos^2 x}{\cos^2 x + \sin^2 x} = \cos^2 x = 1 - \sin^2 x
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    (Original post by nwmyname)
    Here is what I have done so far.

    1 - cos2x
    = 1 - 1/sec2x
    = 1 - 1/1+tan2x

    = (1 + tan2x - 1) / 1+tan2x
    = tan2x / 1 + tan2x

    = tan2x * 1/(1+tan2x)
    = (sec2x - 1) * 1/(1+tan2x)

    How further can I go?

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    OH NOW THAT MAKES SO MUCH MORE EASY SENSE OMD thanks.
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    (Original post by Zacken)
    Overly long winded, but okay. Now notice that:

    \displaystyle \frac{1}{1 + \tan^2 x} = \frac{1}{1 + \frac{\sin^2 x}{\cos^2 x}} = \frac{\cos^2 x}{\cos^2 x + \sin^2 x} = \cos^2 x = 1 - \sin^2 x
    Lemme try and do it now.

    1 - cos2x
    = 1 - (1/sec2x)
    = 1 - (1/1+tan2x)
    = 1+ tan2x -1 / 1 + tan2x
    = tan2x / 1 + tan2x
    = tan2x * (1 / 1 + tan2x)
    = tan2x * (1 / sec2x)
    = tan2x * cos2x
    = (sec2x - 1)(1 - sin2x)

    Very long way of proving it.
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    (Original post by nwmyname)
    OH NOW THAT MAKES SO MUCH MORE EASY SENSE OMD thanks.
    Lol xD


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    (Original post by nwmyname)
    Very long way of proving it.
    Indeed.
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    (Original post by nwmyname)
    I have the following identity to prove.

    From LHS I did
    = 1 - cos2x
    = 1 - 1/cos2x
    = cos2x - 1 / cos2x

    How further can I go to reach the RHS?
    (Original post by Zacken)
    Indeed.
    This is what I did from the get go. Is this a ok way to do it? Would this have been what you did Zack?



    Annoyed I couldn't do basic stuff like this in the exam. :argh: ... time to look forward to self-teaching
    FP2 in 2 1.5 weeks
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    (Original post by XxKingSniprxX)
    This is what I did from the get go. Is this a ok way to do it? Would this have been what you did Zack?

    What you've shown is that the given identity implies a true statement, but you can't do that as you've started by assuming the statement that you're required to prove. Instead, either start with one side of the identity and show that it equals the other side, or start with both sides but simplify each separately and show that they equal the same thing.
 
 
 
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