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Math Help - Where did I go wrong in solving this trigonometry equation?

  1. #1
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    Where did I go wrong in solving this trigonometry equation?

    1 + sec Θ = tan Θ (all solutions on [0,2pi))

    1 + sec Θ = sin/cos Θ (tan is the same as sin/cos)

    cos Θ + 1 = sin Θ (Multiplied both sides by cos, cos*sec = 1)

    cos^2 Θ + 2cos Θ + 1 = sin^2 Θ (Squared both sides)

    cos^2 Θ + 2cos Θ + 1 = (1 - cos^2 Θ) (Pythagorean identity for sin^2 Θ)

    cos^2 Θ + 2cos Θ = - cos^2 Θ (Took away 1 from both sides)

    2cos^2 Θ + 2cos Θ = 0 (Added cos^2 Θ to both sides)

    2cos^2 Θ = -2cos Θ (Moved 2cos Θ to the right side)

    cos^2 Θ = -cos Θ (Divided both sides by 2)

    cos^2 Θ/cos Θ = -1 (Divide both sides by cos Θ)

    cos Θ = -1 (Rules of exponents)

    I must've done a bunch of unnecessary stuff, but there's so many ways to write trigonometry expressions. I skipped to the answer page of my workbook and it said the answer was 0. I guess I'm wrong because arccos(-1) does not equal 0. Can someone help me out?
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  2. #2
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    Hi!
    First of all, before you start solving any equation, you have to specify the set, within which you may search for solutions (the domain). In this case, it is said to be (0, 2pi) but you've got a tangent in this equation, and it's undefined for some values from the set (pi).
    So the domain is (0, 2pi) \ {pi} (excluding pi).

    Everything seems to be correct until this line;
    2cos^2 Θ + 2cos Θ = 0
    Why don't you extract 2cosΘ from brackets and divide by 2 like this:
    cosΘ(cosΘ + 1) = 0
    Now at least one of the factors must equal 0, so either cosΘ=0 or cosΘ=-1. But if cosΘ=-1 then Θ=pi, but pi was excluded from the domain. So the answer is, cosΘ=0



    //Edit: Looking at the post below mine, I found a mistake in my solution... Of course you must also have the secant defined, so cosΘ actually can't equal 0 because it's the denominator if you consider secΘ=1/cosΘ. Sorry for misleading you.
    Last edited by pinkparrot; April 13th 2009 at 12:08 AM. Reason: a mistake found
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  3. #3
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    Quote Originally Posted by Phire View Post
    1 + sec Θ = tan Θ (all solutions on [0,2pi))

    [snip]

    I skipped to the answer page of my workbook and it said the answer was 0. I guess I'm wrong because arccos(-1) does not equal 0. Can someone help me out?
    If θ = 0 then tan θ = 0 and sec θ = 1. So 1 + sec θ = 2 ≠ tan θ. Therefore either the answer in the workbook is wrong (and your answer is right), or you copied the question down wrongly. Maybe the question should have read 1 sec θ = tan θ?
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