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Trig Identities

  1. Jan 5, 2014 #1
    Hello,

    i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

    1. The problem statement, all variables and given/known data

    Solve the following equations in the range -180°≤ θ ≤ 180°

    tanθ + cotθ = 2

    2. Relevant equations



    3. The attempt at a solution

    tanθ + cotθ = 2

    [itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

    [itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

    [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

    [itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

    Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

    then

    [itex]2cos\theta sin\theta[/itex] = 1

    (the above is where I think i've done something careless)

    I now have

    [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

    I don't know what to do with it really... I tried saying [itex]sin^{2}\theta[/itex] = [itex] 1 - cos^{2}\theta[/itex] but it doesn't really get me anywhere.


    Is what I have done not correct?

    EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.
     
  4. Jan 5, 2014 #3
    Ahhh, I was focused on the other side of the equation!

    2cosθsinθ = sin2θ

    sin2θ = 1

    2θ = 90

    θ = 45

    from the curve I can see where -135 comes from also.
     
  5. Jan 5, 2014 #4

    Mark44

    Staff: Mentor

    It's quicker to replace cot(θ) by 1/tan(θ)

    tanθ + cotθ = 2
    tanθ + 1/tanθ = 2

    Multiply by tanθ to get
    tan2θ + 1 = 2tanθ
    tan2θ - 2tanθ + 1 = 0
    This equation can be factored to solve for tanθ.
     
  6. Jan 5, 2014 #5

    Ray Vickson

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    The quantity ##t = \tan(\theta)## must satisfy the equation
    [tex] t + \frac{1}{t} = 2,[/tex]
    whose unique (real) solution is ##t = 1##. This is easy to see and prove. For any ##A,B > 0## the arithmetic-geometric inequality says
    [tex] \frac{A+B}{2} \geq \sqrt{AB},[/tex]
    with equality only when ##A = B##. Apply this to ##A = t, B = 1/t, ## to conclude that for ##t > 0## we have ##t + 1/t \geq 2,## with equality only when ##t = 1##. For ##t < 0## it similarly follows that ##t + 1/t \leq -2,## so negative values of ##t## won't work.

    So, the unique root is ##\tan(\theta) = 1##.
     
  7. Jan 5, 2014 #6
    This is how I solved it after struggling with the method outlined in the OP.

    Much simpler, but it is cool to see the many ways of solving these problems.
     
  8. Jan 13, 2014 #7
    Yes, you've said that [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

    and [itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

    Now, substitute the [itex]2cos\theta sin\theta[/itex] into the numerator of [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

    This will give you [itex]\frac{2cos\theta sin\theta}{cos\theta sin\theta}[/itex] = 2

    which simplifies into 2 = 2
     
  9. Jan 13, 2014 #8

    ehild

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    Homework Helper
    Gold Member

    sin(2θ)= 1 when 2θ =90 °+ k 360 °, k=0, ±1 ±2.... . That means θ=45 ° ± k 180 °. Which angles are in the range [-180°, 180°]?

    ehild
     
  10. Jan 14, 2014 #9

    Mark44

    Staff: Mentor

    So? The idea is to show that tanθ + cotθ = 2, not that 2 = 2, which is obviously true.
     
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