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Math Help - trig identities: eliminating θ

  1. #1
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    trig identities: eliminating θ

    Hello. Thanks for any help in advance.

    just having trouble with the method of eliminating θ from the following pairs of equations:
    a. x = 2 cosθ, y = 2 sin θ

    b. x = b tan, y = a sec θ
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  2. #2
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    Quote Originally Posted by shimmerblade View Post
    Hello. Thanks for any help in advance.

    just having trouble with the method of eliminating θ from the following pairs of equations:
    a. x = 2 cosθ, y = 2 sin θ

    b. x = b tan, y = a sec θ
    The key is to use the many forms of the Pythagorean Identity - in this case \sin^2{\theta} + \cos^2{\theta} = 1 and \tan^2{\theta} + 1 = \sec^2{\theta}.


    (a) x = 2\cos{\theta}, y = 2\sin{\theta}

    x^2 = 4\cos^2{\theta}, y^2 = 4\sin^2{\theta}


    x^2 + y^2 = 4\cos^2{\theta} + 4\sin^2{\theta}

    x^2 + y^2 = 4(\cos^2{\theta} + \sin^2{\theta})

    x^2 + y^2 = 4\cdot 1

    x^2 + y^2 = 4.


    Can you follow a similar process for (b)?
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    for b. x = b tan θ, y = a sec θ would it be something like

    (b tan θ)^2 + 1 = (a sec θ)^2

    b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

    b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

    b^2 x^2 + a^2 b^2 = a^2 y^2

    not too sure....=\
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  4. #4
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    Quote Originally Posted by shimmerblade View Post
    for b. x = b tan θ, y = a sec θ would it be something like

    (b tan θ)^2 + 1 = (a sec θ)^2

    b^2 tan^2 θ + b^2 a^2 = a^2 sec^2 θ

    b^2 (tan^2 θ + a^2)^2 = a^2 (sec^2 θ)

    b^2 x^2 + a^2 b^2 = a^2 y^2

    not too sure....=\
    You're on the right track. Remember that \tan^2{\theta} + 1 = \sec^2{\theta}, so \sec^2{\theta} - \tan^2{\theta} = 1.


    If x = b\tan{\theta}, y = a\sec{\theta}

    then x^2 = b^2\tan^2{\theta}, y^2 = a^2\sec^2{\theta}

    a^2x^2 = a^2b^2\tan^2{\theta}, b^2y^2 = a^2b^2\sec^2{\theta}.



    a^2b^2\sec^2{\theta} - a^2b^2\tan^2{\theta} = b^2y^2 - a^2x^2

    a^2b^2(\sec^2{\theta} - \tan^2{\theta}) = b^2y^2 - a^2x^2

    a^2b^2\cdot 1 = b^2y^2 - a^2x^2

    a^2b^2 = b^2y^2 - a^2x^2

    1 = \frac{b^2y^2 - a^2x^2}{a^2b^2}

    \frac{b^2y^2}{a^2b^2} - \frac{a^2x^2}{a^2b^2} = 1

    \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1.
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  5. #5
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    ok thanks for your help.
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