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Dama28

  • 2 years ago

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π) : tan θ + cot θ = 4 sin 2θ

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  1. Callisto
    • 2 years ago
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    Hmm... \[tan θ + cot θ\]\[= \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ} =...?\]

  2. Dama28
    • 2 years ago
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    I did that and after that I got: \[\frac{ \sin \theta ^{2}+\cos \theta^2 }{ \cos \theta \sin \theta }=4\sin^2\]

  3. Dama28
    • 2 years ago
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    after that: \[\frac{ 1 }{ \cos \theta \sin \theta} =4\sin2\theta \]

  4. Callisto
    • 2 years ago
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    2cosθsinθ = sin2θ Divide both sides by 2 cosθsinθ = ...?

  5. Dama28
    • 2 years ago
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    \[\frac{ 1 }{ \sin2\Theta }=2\sin2\Theta \]

  6. Callisto
    • 2 years ago
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    Multiply both sides by sin2θ, what do you get?

  7. Dama28
    • 2 years ago
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    wait! would it be \[1=2\sin ^{2}2\Theta \] or \[1= 2\sin^2 4\theta\]

  8. Callisto
    • 2 years ago
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    The first one.

  9. Dama28
    • 2 years ago
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    wait, haha I'm sorry!...

  10. Dama28
    • 2 years ago
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    its now \[\frac{ 1 }{ 2 }=\sin^2 \theta\]

  11. Callisto
    • 2 years ago
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    Next: Use cos2x = 1-2(sinx)^2

  12. Dama28
    • 2 years ago
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    so I take the square root?? and get sin(\[\sin\frac{ \sqrt{2} }{ 2 }=2\theta \]

  13. Dama28
    • 2 years ago
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    riiiight! I see it.

  14. Callisto
    • 2 years ago
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    \[cos2x = 1-2sin^2x\]\[2sin^2x = 1-cos2x\]Replace x by 2θ. Life isn't difficult then.

  15. Callisto
    • 2 years ago
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    It should be\[\sin^{-1}\frac{ \sqrt{2} }{ 2 }=2\theta\]

  16. Dama28
    • 2 years ago
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    i get \[2\Theta = \frac{ \pi }{ 2} \] and \[\frac{ 3\pi }{ 2 }\]

  17. Callisto
    • 2 years ago
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    Are you sure...???

  18. Callisto
    • 2 years ago
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    \[sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

  19. hkm91
    • 2 years ago
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    2theta =pi/4, 3pi/4

  20. Dama28
    • 2 years ago
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    so, \[\Theta=\frac{ \pi }{ 8 } and \frac{ 3\pi }{ 8 }\] ??

  21. hkm91
    • 2 years ago
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    yupp

  22. Dama28
    • 2 years ago
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    now, how do I find the the equation in the interval [0, 2π)??

  23. Callisto
    • 2 years ago
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    Hmm... Something's wrong there. \[1=2\sin ^{2}2\theta\]\[sin^2 2\theta = \frac{1}{2}\]\[sin2\theta = \pm \sqrt{\frac{1}{2}}\]So, \[sin2\theta = \sqrt{\frac{1}{2}}\]or \[sin2\theta = -\sqrt{\frac{1}{2}}\]

  24. Dama28
    • 2 years ago
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    i noticed something, since the square root was taken to take out the squared sine....then the result needs to be +- \[\pm \frac{ \sqrt{2} }{ 2 }\]

  25. Callisto
    • 2 years ago
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    :)

  26. Dama28
    • 2 years ago
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    wow! at the same time

  27. Callisto
    • 2 years ago
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    To avoid taking square root, I prefer using my method: \[1=2\sin ^{2}2\theta\]\[1=1-cos4\theta\]\[cos4\theta =0\]Then. solve theta.

  28. Dama28
    • 2 years ago
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    ohhhh ok

  29. Callisto
    • 2 years ago
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    But it's up to you :)

  30. Dama28
    • 2 years ago
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    ok, so when I solve for all the solutions...is it?: \[\frac{ \pi }{ 8 }, \frac{ 3\pi }{ 8 },\frac{ 5\pi }{ 8 }, \frac{ 7\pi }{8 }\]

  31. EulerGroupie
    • 2 years ago
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    From zero to pi, yes... keep going to 2pi... 4 more solutions.

  32. EulerGroupie
    • 2 years ago
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    The interval described is 0 to 2pi... so all of the answers that exist between 0 and 2pi are what you are looking for.

  33. Callisto
    • 2 years ago
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    If you want to get the next 4 solutions, use: 2pi - solution you've got. Then you'll have 4 more. Eg: 2pi - (pi/8)

  34. Callisto
    • 2 years ago
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    What have you put?

  35. Dama28
    • 2 years ago
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    pi/8, 3pi/8 my first try and pi/8, 3pi/8, 5pi/8, 7pi/8 the second try. I have one more try :S

  36. EulerGroupie
    • 2 years ago
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    pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi/8 are all of the solutions between zero and 2pi. I wish I could show you my pretty graph on my calculator. :)

  37. EulerGroupie
    • 2 years ago
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    I don't like to just give answers, but you are working very hard at this. ;)

  38. Callisto
    • 2 years ago
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    First four: pi/8, 3pi/8, 5pi/8, 7pi/8 Next four: 2pi - pi/8 = 15pi/8 2pi - 3pi/8 = 13pi/8 2pi - 5pi/8 =... 2pi - 7pi/8 =... I'm sorry :(

  39. EulerGroupie
    • 2 years ago
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    My only concern is... was that interval stated as [0, 2pi] or [0, 2pi) ? You have said each through the course of this chat.

  40. Dama28
    • 2 years ago
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    hahaha thanks. the reason why I didn't keep going is because in the previous problems I had, if it was divided by 3 theta I would only give 6 answers. If if was divided by 4theta, then 8 answers, so I would assume that in this excercise that's divided by 2theta I would only give 4 answers.

  41. Dama28
    • 2 years ago
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    its [0, 2pi)

  42. EulerGroupie
    • 2 years ago
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    ok... I stand by my answer... and @Callisto 's

  43. Dama28
    • 2 years ago
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    ok. hahahah thank you so sooooo much! and I'm sorry for the trouble

  44. EulerGroupie
    • 2 years ago
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    Oh, this stuff is fun... no troubles. We do it because we love it... especially when people are really working and trying instead of fishing for easy answers.

  45. Dama28
    • 2 years ago
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    ok. thanks!

  46. Callisto
    • 2 years ago
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    Agree with @EulerGroupie 's saying :)

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