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Trigonometry problem

  1. May 18, 2016 #1
    1. The problem statement, all variables and given/known data

    If ##cos^3x+\frac{1}{cos^3x} = 0##, then sin 2x equals ...

    A. -1
    B. -√3
    C. -√2
    D. 1
    E. 2

    2. Relevant equations
    Basic trigonometry

    3. The attempt at a solution

    Let y = cos^3 x

    y+1/y = 0
    y^2 + 1 = 0
    y = √(-1)

    Then, I don't know what to do.. Help please..

     
  2. jcsd
  3. May 18, 2016 #2

    Math_QED

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    I think you can solve this one for cosx. Make the expression into one fraction and remember a/b = 0 <=> a = 0 when b =/= 0.
     
  4. May 18, 2016 #3

    ##\frac{cos^6x+1}{cos^3x}=0##
    ##cos^6x = -1##
    ##cos x =\sqrt[6]{-1}##
    ##sin 2x = 2 sin x cos x \\
    sin 2x = 2 \sqrt{1-\sqrt[3]{-1}}\sqrt[6]{-1}##

    How to solve it????
     
  5. May 18, 2016 #4

    Math_QED

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    Well, in the context of real numbers, there does not seem to be an x that satisfies the equation.
     
  6. May 18, 2016 #5
    So, no option satisfies??
     
  7. May 18, 2016 #6

    blue_leaf77

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    Then ##x## must be complex. From the original equation, out of the two possibilities you can choose ##\cos^3 x = i = e^{i\pi/2}##. From this, find ##\cos^2 x## and ##\cos x##, then ##\sin x##.
     
  8. May 18, 2016 #7
    It seems a bit complicated to take the cubic root from exponential form.
    Is there any easier way??
    During test, calculator is not allowed.. So I try not using calculator since this is my preparation
     
  9. May 18, 2016 #8

    blue_leaf77

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    No, it's fairly easy actually.
    From ##\cos^3 x = e^{i\pi/2}##, find ##\cos x ## in terms of complex exponential.
     
  10. May 18, 2016 #9
    ##\cos x = e^{i\pi/6} \\
    sin x = \sqrt{1-e^{i\pi/3}} \\
    sin 2x = 2 sin\ x\ cos\ x\ = 2\sqrt{1-e^{i\pi/3}} e^{i\pi/6}##

    Now, it's complicated.
    How to solve it?
     
  11. May 18, 2016 #10

    blue_leaf77

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    Not that complicated if you have put some more creativity and trials, now in
    $$
    sin x = \sqrt{1-e^{i\pi/3}} \\
    $$
    express ##1-e^{i\pi/3}## in ##x+iy## form.
     
  12. May 18, 2016 #11
    Hmmm... I'm new to complex number... and, I don't know how to convert ##1-e^{i\pi/3}## to rectangular form since there is a constant (in this case, 1)...
    Please help..
     
  13. May 18, 2016 #12

    blue_leaf77

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    In rectangular form of a complex number ##x+iy##, both ##x## and ##y## are real and every complex number can be transformed to this form. To do what you need to do, transform ##e^{i\pi/3}## into rectangular form using Euler formula.
     
  14. May 18, 2016 #13
    Alright,
    so it becomes like this
    ##1-e^{i\pi/3} = 1 - cos \frac{\pi}{3} + i\ sin \frac{\pi}{3} = \frac{1}{2} + \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1+\sqrt{3}i) \\
    sin\ x = \sqrt{\frac{1}{2} (1+\sqrt{3}i)} ##

    and I convert cos x to rectangular form,

    ## cos\ x = \frac{1}{2}\sqrt{3}+\frac{1}{2}i##

    ##sin\ 2x = 2\ sin\ x\ cos\ x\ = 2 (\sqrt{\frac{1}{2} (1+\sqrt{3}i)} )( \frac{1}{2}\sqrt{3}+\frac{1}{2}i)##

    Still seems complicated :(
     
  15. May 18, 2016 #14

    blue_leaf77

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    You make a mistake ##e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}##. After incorporating this correction, you will get ##\sin^2 x## in rectangular form, then cast it back into polar form. Take its square root to obtain ##\sin x##.
     
  16. May 18, 2016 #15

    SteamKing

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    After looking at all the complex algebra and trig which has been expended on solving this problem in the posts above, you can eliminate some of the answers A. - E., at least in terms of real solutions for x.

    The problem statement wants to know what the value of sin (2x) is, and like all sine functions, its amplitude is bound to the interval [-1, 1], so that means that any choices whose value lies outside this range cannot be correct. The remaining choices can be easily checked to see if they can satisfy the original cosine equation.

    Given the comments made so far in this thread, I can't help but wonder if something has been left out of the original problem statement.
     
  17. May 18, 2016 #16

    blue_leaf77

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    If the argument is complex, then it can be bigger than unity. For example ##\cos (iy) = \cosh y \geq 1##.
     
  18. May 18, 2016 #17
    I forgot to put parentheses
    ##1-e^{i\pi/3} = 1 - (cos \frac{\pi}{3} + i\ sin \frac{\pi}{3}) = \frac{1}{2} - \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1-\sqrt{3}i) \\
    sin\ x = \sqrt{\frac{1}{2} (1-\sqrt{3}i)}##
    ##1-e^{i\pi/3} = 1 - (cos \frac{\pi}{3} + i\ sin \frac{\pi}{3}) = \frac{1}{2} - \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1-\sqrt{3}i) \\
    sin\ x = \sqrt{\frac{1}{2} (1-\sqrt{3}i)} \\
    sin\ 2x = 2\ sin\ x\ cos\ x\ = 2 (\sqrt{\frac{1}{2} (1-\sqrt{3}i)} )( \frac{1}{2}\sqrt{3}+\frac{1}{2}i)##

    Stuck again.
     
  19. May 18, 2016 #18

    blue_leaf77

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    Like I said, cast ##\frac{1}{2} (1-\sqrt{3}i)## back to polar form.
     
  20. May 18, 2016 #19
    Polar form?? Do you mean something like r (cos alpha + i sin alpha) ?
    ##r = \sqrt{{\frac{1}{2}}^2+({\frac{1}{2}}\sqrt{3})^2} = \sqrt{1} = 1 \\
    \alpha = tan^{-1} (\frac{-\sqrt{3}}{1}) \\
    \alpha = \frac{5}{3}\pi \\
    \frac{1}{2}(1-\sqrt{3}i) = 1 (cos \frac{5}{3}\pi+i sin\frac{5}{3}\pi) \\##

    Taking the root of it?? Why should I make it in polar form??
     
  21. May 18, 2016 #20

    blue_leaf77

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    I mean complex exponential form, ##re^{i\theta}##.
    You don't need to actually but doing so will simplify your calculation.
     
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