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    • Thread Starter
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    Prove that:

    (1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

    I get too many cosines.

    Laura xxxx
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    (Original post by Laura_M)
    Prove that:

    (1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

    I get too many cosines.

    Laura xxxx

    1-tan²θ/1+tan²θ

    1+tan²θ = sec²θ

    => = (1-tan²θ ) x cos²θ

    =cos²θ - tan²θcos²θ

    since tan²θ = sin²θ/cos²θ

    => = cos²θ - cos²θsin²θ/cos²θ

    = cos²θ - sin²θ = cos2θ

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    (Original post by Laura_M)
    Prove that:

    (1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

    I get too many cosines.

    Laura xxxx
    cos^2 x +sin^2 x=1
    1+tan^2 x =sec^2x

    this gives 1-tan^2 x/sec^2 x
    = 1/sec^2 x- (tan^2 x)/sec^2 x
    =cos^2 x- cos^2 x (tan^2 x)
    =cos^2 x -sin^2 x
    =cos 2x
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    please, would people stop making topic titles like 'help please'

    be more specific
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    (Original post by Chiral)
    please, would people stop making topic titles like 'help please'

    be more specific
    people? you mean me?
    be more specific
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    Oooh, pwnage! :O

    But to be fair, Chiral's got a point.

    [And, also, to be fair, you (Laura) are not the only one to make an ambiguous thread. So 'people' in general is true.
    But still, nice pwning. Sorta.]
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    (Original post by KAISER_MOLE)
    1-tan²θ/1+tan²θ

    1+tan²θ = sec²θ

    => = (1-tan²θ ) x cos²θ

    =cos²θ - tan²θcos²θ

    since tan²θ = sin²θ/cos²θ

    => = cos²θ - cos²θsin²θ/cos²θ

    = cos²θ - sin²θ = cos2θ


    How did you get from the 3rd line to the 4th line?
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    (Original post by mockel)
    Oooh, pwnage! :O

    But to be fair, Chiral's got a point.

    [And, also, to be fair, you (Laura) are not the only one to make an ambiguous thread. So 'people' in general is true.
    But still, nice pwning. Sorta.]

    Lol
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    (Original post by Laura_M)
    How did you get from the 3rd line to the 4th line?
    He expanded the brackets:

    cos²x(1-tan²x) = cos²x - cos²x.tan²x

    etc...
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    (Original post by mockel)
    Oooh, pwnage!
    What's 'pwnage' ?
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    (Original post by Gaz031)
    What's 'pwnage' ?
    Ownage. (please say you know what that is.... :rolleyes: )
    • Thread Starter
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    Thank you v. much.
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    http://www.google.com/search?hl=en&l...fine%3A+pwnage

    The second defenition gives a nice history to the word "pwn" or similar.
    "Pwnage" is just a 'branch' of "pwn" (as you can see from the first defenition).
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    (Original post by Laura_M)
    Prove that:

    (1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

    I get too many cosines.

    Laura xxxx
    Alternative solution:

    LHS = (1-tan^2θ)/(1+tan^2θ) = [1 - (sin²θ/cos²θ)]/[1 + (sin²θ/cos²θ)]

    => [cos²θ - sin²θ]/[cos²θ + sin²θ]

    => (cos2θ)/1 = cos2θ = RHS.
 
 
 
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