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    (Original post by Sourestdeeds)
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    Post what you've tried -
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    (Original post by Sourestdeeds)
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    Are you sure this is C3? It seems to require C4 identities (unless non-WJEC boards include those in C3).
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    (Original post by Hydeman)
    Are you sure this is C3? It seems to require C4 identities (unless non-WJEC boards include those in C3).
    Its from the OCR C3 Jan 2013 paper
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    I doubt my attempt is much use! Its just I've not come across a double angle formula which is squared, so not sure how to approach.
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    To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution
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    (Original post by Sourestdeeds)
    To begin with i was barking up the wrong tree and trying to prove it was cos squared instead of sin squared, but later on i tried a different route with this in mind and ended up with the same solution
    Remember that

    \cos^2(a+b) = (\cos(a+b))^2 = (\cos(a)\cos(b) - \sin(a)\sin(b))^2

    so you've missed a term in the second line.
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    ill have a go
    cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x



cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x



 cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x



2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 



2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0

    Continue from here
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    (Original post by SCAR H)
    ill have a go
    cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x



cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x



 cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x



2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 



2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0

    Continue from here
    Its an identity? Surely you start with the left hand side to prove the right hand side?
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    (Original post by Sourestdeeds)
    Its an identity? Surely you start with the left hand side to prove the right hand side?
    Hmm didnt see that

    Works the same way though
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    (Original post by SCAR H)
    ill have a go
    cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = sin^2x



cos^2(x-45) - \frac{1}{2}(cos2a-sin2a) = 1 -cos^2x



 cos^2(x-45) - \frac{1}{2}((sin^2x-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(((1-cos^2x)-cos^2x)-sin2a) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



cos^2(x-45) - \frac{1}{2}(1-2cos^2x-sin2x) = 1 - cos^2x 



2cos^2(x-45) - 1 + 2cos^2x + sin2x = 2- 2cos^2x



2cos^2(x-45) - 3 + 4cos^2x + sin2x = 0 



2cos^2(x-45) - 3 + 4cos^2x + 2(cosx)(sinx) = 0

    Continue from here
    (Original post by Sourestdeeds)
    Its an identity? Surely you start with the left hand side to prove the right hand side?
    Yes, the problem with that kind of thing is that you're already assuming that the identity holds, so it's not a proof.

    You have to start with one side and show that it's equivalent to the other.
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    (Original post by Indeterminate)
    Yes, the problem with that kind of thing is that you're already assuming that the identity holds, so it's not a proof.

    You have to start with one side and show that it's equivalent to the other.
    Yes
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    Did it!

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    Im pretty ill at the moment, making some hilarious mistakes haha. Apparently the LHS is the same as the LHS. Perfect logic
 
 
 

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