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Trigonometry Review: Conflict Answer, Probably Me

  1. Oct 22, 2008 #1
    SOLVED! Thank-you jhicks and Tedjn!

    I am taking a basic calculus course and have some weaknesses when it comes to trigonometry. In this case, it's pure trig.

    1. The problem statement, all variables and given/known data
    The question states: "Find all values of x in the interval [0,2pi] that satisfy the equation sin x = tan x".

    2. Relevant equations
    I do know that: tan x = sin x / cos x

    3. The attempt at a solution
    Since I know that tan x = sin x / cos x, I can rewrite the above equation sin x = tan x to:
    sin x = sin x / cos x
    Rearranging and canceling terms, I get:
    cos x = 1
    So my answer to this problem would be x = 0 and 2pi (in radians, of course), that satisfy the given interval.
    I based this answer on the "Trig Functions of Important Angles" and worked out the multiples of pi that satisfied the equation cos x = 1.

    My problem is (lol, isn't it always) with the answer key. While they agree that 0 and 2pi are correct, they also add pi. Why? When I consider the cosine of pi (in radians), I get -1, not one. Pi doesn't seem to agree with cos x = 1 when x=pi.
    Yet, when I work out sin x = tan x, using x=pi, it works out to 0=0, which is certainly true. How did I miss it? Where'd I go wrong?
     
    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 22, 2008 #2
    Don't ever divide out terms like that. You get rid of roots when that happens. In this case, your modified equation fails to see that 0 = 0/-1
     
  4. Oct 22, 2008 #3
    jhicks... not sure what you're getting at. Here's what I did when I said I "Rearranged and canceled" terms:

    sin x = sin x / cos x (because tan x = sin x / cos x)
    sin x * cos x = sin x (multiply both sides by cos x)
    cos x = sin x / sin x (divide both sides by sin x)
    cos x = 1 (left with... cos x = 1)

    When you say, "Don't ever divide out terms like that. You get rid of roots when that happens. In this case, your modified equation fails to see that 0 = 0/-1", I don't understand. Could you please elaborate?
     
  5. Oct 22, 2008 #4
    You lose solutions when you divide by 0. When you divide both sides by sin x, you are dividing by 0 if x = 0, pi, 2pi, etc., and division by 0 is illegal. Instead, the typical trick for these cases is to factor.

    sin(x)cos(x) = sin(x) => sin(x)cos(x) - sin(x) = 0 => sin(x)(cos(x)-1) = 0. Here, we see clearly that we have solutions whenever sin(x) = 0 or cos(x) - 1 = 0.
     
  6. Oct 23, 2008 #5
    Ohhhh, okay, I think I get what you're saying. Because I cancelled out the sin x term and not just factored it, I lost it; same idea as solving for x -> (x+2)(x+4)=0 ; x=-2 or -4; in this example I divided out the (x+4) and only arrived that x=-2, ignoring the (x+4), which I can't forget about....

    Okay, I think I get it....I felt that little light bulb go off. I know where to come if I run into more trouble.

    Thanks all!
     
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