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Homework Help: Trig problem

  1. Apr 27, 2012 #1
    solve this equation

    after some symbol shunting this simplifies to sin-cos=1

    I believe there is no solution to this problem. Let sin = a and cos = b

    if a-b=1



    but we know that a2+b2 must =1 therefor, no solution. Is this a valid proof?
  2. jcsd
  3. Apr 28, 2012 #2


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    Not at all, a=1 and b=0 solves a-b=1.
  4. Apr 28, 2012 #3
    Well then I am stumped. Are you saying there is a solution? It certainly doesn't look it when I graph it.
  5. Apr 28, 2012 #4
    Of course there is a solution. What's the max value that sin(x) or cos(x) will output? Is there any angle x, such that sin(x) or cos(x) will output 1, and the equation sin(x) - cos(x) = 1 is still satisfied?

    Clearly from that equation, you know that either:
    a.) cos(x) outputs a value >0, and then to satisfy, sin(x) must output a value >1.
  6. Apr 28, 2012 #5


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    You must be graphing it incorrectly then. Think about values of x where sin(x) is positive and cos(x) is negative. Can't it certainly be possible that sin(x)-cos(x)>1 in this case?

    To begin solving this problem, try converting sin(x)-cos(x) into [itex]R\sin(x+\alpha)[/itex] for some constants R and [itex]\alpha[/itex] that you need to determine.
  7. Apr 28, 2012 #6
    Well when you put it like that it's pretty damn obvious :shy:

    Graphing [itex]\frac{1}{sin(x)}[/itex]+[itex]\frac{1}{tan(x)}[/itex] yields holes at y=1 and y=-1. What other way is there to graph it?
  8. Apr 28, 2012 #7


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    What does that graph have to do with anything? In particular, what does tan(x) have to do with anything? You are asking about the equation csc(x)+ cot(x)= 1. At [itex]x= k\pi/2[/itex], csc(x)= 1 and cot(x)= 0. THAT'S what you should be graphing.
  9. Apr 28, 2012 #8
    He was graphing the original equation in order to see where it equals 1.
  10. Apr 28, 2012 #9
    [itex]\frac{1}{tan}[/itex] is cotan, and until my TI-84 magically grows cot, sec and csc buttons that's how I'll have to graph it.
  11. Apr 28, 2012 #10
    If you know half angle formulas, then your original equation reduces to:
    \csc x + \cot x = \frac{1 + \cos x}{\sin x} = \frac{2 \, \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \cot \frac{x}{2}, \ \cos \frac{x}{2} \neq 0
    You may check that when:
    \cos {\frac{x}{2} } = 0 \Leftrightarrow \frac{x}{2} = \frac{\pi}{2} + n \, \pi \Rightarrow x = (2 n + 1) \pi
    \sin x = \sin \left[ (2 n + 1) \pi \right]
    so there is no loss of solutions by canceling with [itex]\cos ( x/2 )[/itex]. On, the other hand, if you multiply the equation by [itex] \sin x[/itex], and convert it as you did to:
    \sin {x} - \cos{x} = 1
    you may see that [itex]x = (2 n + 1) \pi[/itex] are solutions of this equation, whereas they are not for the original equation.
  12. Apr 28, 2012 #11
    I can tentatively follow your logic, but I'm still not clear on what the solution is. Is the answer x= [itex]\frac{∏}{2}[/itex] [0,2∏) or no solution? If my simplification generated an invalid answer then surely yours did as well because neither works in the original equation.
  13. Apr 28, 2012 #12
    \cot \frac{x}{2} = 1
    has the solution:
    \frac{x}{2} = \frac{\pi}{4} + n \pi \Rightarrow x = \frac{\pi}{2} (4 n + 1)

    The set of solutions:
    (2 n + 1) \pi = (4 n + 2) \frac{\pi}{2}
    is not of the above form, so do not need to exclude any of them.
  14. Apr 28, 2012 #13
    That doesn't answer my question.
  15. Apr 28, 2012 #14
    What was your question?
    Also, I'm not sure about x=∏/2 [0,2∏) a couple posts ago.

    Dickfore did get the correct general solution of [itex]x=\frac{\pi}{2}(4n+1)[/itex]

    I also managed to rewrite the original equation as sinx - cosx = 1, but this has one set of solution, x = 2πn + π, which doesn't work for the original equation.
  16. Apr 29, 2012 #15
    Again, my question is..

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