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    Hey guys I need proving the following identity, where t = theta. :

    1-2sin^2 t / cos t + sin t = cos t - sin t.
    How can start to use identities on the left hand side?

    Many thanks
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    You have \cos t on the denominator of the fraction, and no other fractions. This implies that you want to somehow write the \sin^2 t on the top of the fraction in terms of \cos \theta (or \cos^2 \theta)... what identities do you know that link these things together?

    EDIT: I'm not convinced that the identity you've given is actually valid. Are you sure you've copied it correctly?
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    start from the RHS and go to the LHS. Do the RHS x (cost + sint) / (cost + sint) and go from there
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    (Original post by nuodai)
    You have \cos t on the denominator of the fraction, and no other fractions. This implies that you want to somehow write the \sin^2 t on the top of the fraction in terms of \cos \theta (or \cos^2 \theta)... what identities do you know that link these things together?

    EDIT: I'm not convinced that the identity you've given is actually valid. Are you sure you've copied it correctly?
    Yes that's right, and I know that sin^2 t = 1 - cos^2 t, so then my fraction would be: 1 - 2(1-cos^2 t) / cos t + sin t?

    And yes I believe I have copied it correctly, just to state again it says:

    1 - 2sin^2 t / cos t + sin t = cos t - sin t (note the divide sign represents that like a fraction)
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    (Original post by adil12)
    Yes that's right, and I know that sin^2 t = 1 - cos^2 t, so then my fraction would be: 1 - 2(1-cos^2 t) / cos t + sin t?

    And yes I believe I have copied it correctly, just to state again it says:

    1 - 2sin^2 t / cos t + sin t = cos t - sin t (note the divide sign represents that like a fraction)
    I just realised your fraction is supposed to be \dfrac{1-2\sin^2 t}{\cos t + \sin t} and not \dfrac{1-2\sin^2 t}{\cos t}+\sin t like I thought... it's a better idea to use brackets to make it more clear.

    Your best bet is to write 1-2\sin^2 t as (1-\sin^2 t)-\sin^2 t. How else can you write 1-\sin^2 t? Then factorise the numerator of the fraction and cancel common factors. [Hint: difference of two squares.]
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    (Original post by nuodai)
    I just realised your fraction is supposed to be \dfrac{1-2\sin^2 t}{\cos t + \sin t} and not \dfrac{1-2\sin^2 t}{\cos t}+\sin t like I thought... it's a better idea to use brackets to make it more clear.

    Your best bet is to write 1-2\sin^2 t as (1-\sin^2 t)-\sin^2 t. How else can you write 1-\sin^2 t? Then factorise the numerator of the fraction and cancel common factors. [Hint: difference of two squares.]
    Thanks a lot! Repped
 
 
 
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