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Math Help - Trigonometric identities

  1. #1
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    Trigonometric identities

    Please help with any of the following..

    Prove these identities

    cos2x = (1 - tan^2 x)/(1 +tan^2 x)


    cos2x = cos^4 x - sin^4 x


    cosec2x + cot2x = cot x


    sin2x = 2tanx / (1 + tan^2 x)


    tan x = sqrt ((1 - cos2x) / (1 + cos2x))


    Thankyou for your time and help
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  2. #2
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    Quote Originally Posted by BeckyDWood View Post
    Please help with any of the following..

    Prove these identities

    cos2x = (1 - tan^2 x)/(1 +tan^2 x)

    note that 1 + tan^2 x = sec^2 x


    cos2x = cos^4 x - sin^4 x

    factor the right side


    cosec2x + cot2x = cot x

    left side ...

    1/sin(2x) + cos(2x)/sin(2x)

    [1 + cos(2x)]/sin(2x)

    now use cos(2x) = 2cos^2 x - 1 and sin(2x) = 2sinxcosx




    sin2x = 2tanx / (1 + tan^2 x)

    same hint as the first


    tan x = sqrt ((1 - cos2x) / (1 + cos2x))

    on the right side, try these substitutions ...

    numerator ... cos(2x) = 1 - 2sin^2 x

    denominator ... cos(2x) = 2cos^2 x - 1

    ...
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  3. #3
    Senior Member pacman's Avatar
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    use this identity: (cos^2 x + sin^2 x = 1)

    i) cos2x = cos^4 x - sin^4 x, use factoring by difference of 2 squares;

    RHS = (cos^2 x - sin^2 x)(cos^2 x + sin^2 x) = cos^2 x - sin^2 x = cos 2x = LHS


    ii) csc 2x + cot 2x = cot x,

    use: csc y = 1/sin y & cot y = cos y/sin y;

    LHS = (1/(sin 2x) + (cos 2x)/(sin 2x) = (1 + cos 2x)/(sin 2x)

    use these identities: (2cos^2 x = 1 + cos 2x) & (sin 2x = 2sin x cos x)

    LHS = (2cos^2 x)/(sin 2x)

    LHS = 2(cos x)(cos x)/(2 sin x cos x)

    LHS = 2(cos x)/(2 sin x)

    LHS = (cos x)/(sin x)

    LHS = cot x

    LHS = RHS
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  4. #4
    Senior Member pacman's Avatar
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    RHS = sin2x = 2tanx / (1 + tan^2 x) = (2tan x)/(sec^2 x) = (2sin x/cos x)/[(1/cos x)(cos x)] = (2 sin x / cos x)(cos x)(cos x)

    = (2 sin x cos x) = sin 2x = LHS
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