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anonymous

  • one year ago

find the exact value of cos75 degrees

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  1. anonymous
    • one year ago
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    Use a half angle identity (150/2 = 75) or sum and difference identity (30 + 45) = 75 half angle (Θ = 150°): \[\cos \frac{ \theta }{ 2 }=\sqrt{\frac{ 1+\cos \theta }{ 2 }}\] sum and difference (u = 30°, v = 45°) \[\cos (u + v) = \cos u \cos v - \sin u \sin v\]

  2. anonymous
    • one year ago
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    That is just as confusing as the notes

  3. anonymous
    • one year ago
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    this is the example I have in the notes

  4. anonymous
    • one year ago
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    ok so it looks like they used a sum and difference identity: \(cos(u+v)= \cos u \cos v− \sin u \sin v\) The first thing you have to do is find 2 numbers on your unit circle that add up to 75. That's where the 30 and 45 come from. Then let u be 30 and let v be 45 and we're going to plug them in \(cos 75°=cos(30°+45°) With me so far?

  5. anonymous
    • one year ago
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    so far, yes lol

  6. anonymous
    • one year ago
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    I have one as a tab, not printed out

  7. anonymous
    • one year ago
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    that's good enough.

  8. anonymous
    • one year ago
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    So this is our identity cos (u + v) = cos u cos v - sin u sin v Plug in the 30 and 45 cos (30° + 45°) = cos 30° cos 45° - sin 30° sin 45° Now use the unit circle to find the 4 values in the identity: cos 30° cos 45° sin 30° sin 45°

  9. anonymous
    • one year ago
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    why do you subtract it by sin ?

  10. anonymous
    • one year ago
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    your formula is different than the one in my notes :/

  11. anonymous
    • one year ago
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    that's just the identity. Here's a list of the identities. Look under sum and difference http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

  12. anonymous
    • one year ago
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    so how do you know which one to use ?

  13. anonymous
    • one year ago
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    which sign to use, or which identity?

  14. anonymous
    • one year ago
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    identity

  15. anonymous
    • one year ago
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    It depends on what you're trying to find. For this one you can use multiple identities. I started off wanting to use a half angle because then all you'd have to do is find cos 150° and plug it into the formula. You could also have use the difference identity for cosine because 120 - 45 = 75. You're just looking for any combination that works. They'll all give you the same answer

  16. anonymous
    • one year ago
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    so pretty much you can us any, just depending if you're adding or subtracting ?

  17. anonymous
    • one year ago
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    yeah as long as it's the correct trig function and the numbers add/subtract right

  18. anonymous
    • one year ago
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    okay , I understand that part so far

  19. anonymous
    • one year ago
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    cos (30° + 45°) = cos 30° cos 45° - sin 30° sin 45° ok so now we plug in the numbers from the unit circle \[\cos 75°=\frac{ \sqrt3 }{ 2 }\times \frac{ \sqrt2 }{ 2 }-\frac{ 1 }{ 2 }\times \frac{ \sqrt2 }{ 2 }\]

  20. anonymous
    • one year ago
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    \[\cos 75°=\frac{ \sqrt6 }{ 4 }-\frac{ \sqrt2 }{ 4 }\] \[\cos 75°=\frac{ \sqrt6 -\sqrt2 }{ 4 }\]

  21. anonymous
    • one year ago
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    where did the 1/2 go

  22. anonymous
    • one year ago
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    jk

  23. anonymous
    • one year ago
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    you're multiplying xD

  24. anonymous
    • one year ago
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    yeah :)

  25. anonymous
    • one year ago
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    I don't remember how to add and subtract radicals. I always mix it up

  26. anonymous
    • one year ago
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    the numbers under the radicals have to be the same to add or subtract, so this can't be simplified any further

  27. anonymous
    • one year ago
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    so it'd have to be \[\sqrt{4} - \sqrt{4} \] to subtract ?

  28. anonymous
    • one year ago
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    right

  29. anonymous
    • one year ago
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    but wouldn't that just always equal 0 ?

  30. anonymous
    • one year ago
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    yes

  31. anonymous
    • one year ago
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    but if you had something like this, just subtract the "coefficients" \[5\sqrt6-3\sqrt6=2\sqrt6\]

  32. anonymous
    • one year ago
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    oooh okay, it's coming back lol. It's been a longer summer

  33. anonymous
    • one year ago
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    haha everyone's saying that

  34. anonymous
    • one year ago
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    I've been taking summer school for algebra 2 and trigonometry was thrown into algebra 2, which makes no sense to me

  35. anonymous
    • one year ago
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    yeah that doesn't make sense. This is more advanced trig so it really shouldn't be mixed in there

  36. anonymous
    • one year ago
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    I passed both semester of algebra 2 with a D, that being an accelerated class . I'm taking regular algebra 2 for summer school . Idk how the state thinks this is possible for an algebra 2 class lol

  37. anonymous
    • one year ago
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    That sounds like GA. I've tutored here since 2009 and the curriculum has changed 4 times. They keep moving stuff from one class to the next. It's ridiculous

  38. anonymous
    • one year ago
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    Common core is just awful . This year is the first year for common core to be fully enforced by math . Trig is introduced in accelerated algebra 2 when I was taking the class , but it wasn't so in depth

  39. anonymous
    • one year ago
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    @peachpi could you help me with tan105 degrees ?

  40. anonymous
    • one year ago
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    105 = 150 - 45

  41. anonymous
    • one year ago
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    \[\tan(u-v)=\frac{ \tan u - \tan v }{ 1+\tan u \tan v }\] \[\tan(150°-45°)=\frac{ \tan 150° - \tan 45° }{ 1+\tan 150° \tan 45° }\]

  42. anonymous
    • one year ago
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    yeah?

  43. anonymous
    • one year ago
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    to find tan , you have to divide sin by cos right ?

  44. anonymous
    • one year ago
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    yes

  45. anonymous
    • one year ago
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    and then you multiply the denominator out to only have a numerator ?

  46. anonymous
    • one year ago
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    right

  47. anonymous
    • one year ago
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    okay , ima try it and see what I come up with

  48. anonymous
    • one year ago
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    ok

  49. anonymous
    • one year ago
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    you divide by the reciprocal right ?

  50. anonymous
    • one year ago
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    multiply *

  51. anonymous
    • one year ago
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    yes

  52. anonymous
    • one year ago
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    so for tan150 , I got \[-\frac{ 2 }{ 2\sqrt{3} }\] is that right ?

  53. anonymous
    • one year ago
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    uh , no . where did that come from ?

  54. anonymous
    • one year ago
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    oh, wait never mind. I misread what you had. That is correct for tan 150. I read tan 105

  55. anonymous
    • one year ago
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    oh no lol. I'm doing it step by step

  56. anonymous
    • one year ago
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    so then tan45 is 1 ?

  57. anonymous
    • one year ago
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    yes

  58. anonymous
    • one year ago
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    and to substitute , I got \[\frac{ - \frac{ 2 }{ 2\sqrt{3} } - 1 }{ 1+(-\frac{ 2 }{ 2\sqrt{3} }) \times 1}\]

  59. anonymous
    • one year ago
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    Yes that's right

  60. anonymous
    • one year ago
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    how do you factor that lol

  61. anonymous
    • one year ago
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    start with simplifying: -2/(2√3) = -1/√3 Then multiply by √3/√3 to clear denominators |dw:1440272933438:dw|

  62. anonymous
    • one year ago
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    oooh okay so you have to get rid of the denominator for tan150 fraction ?

  63. anonymous
    • one year ago
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    for this one yeah because you can't have the radical in the denominator

  64. anonymous
    • one year ago
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    thank you!

  65. anonymous
    • one year ago
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    you're welcome

  66. anonymous
    • one year ago
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    to rationalize the denominator , how do you do that ? ... xD

  67. anonymous
    • one year ago
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    jk , I looked at your answer and I figured it out

  68. anonymous
    • one year ago
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    multiply by the conjugate

  69. anonymous
    • one year ago
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    oh ok. yeah foil will do it

  70. anonymous
    • one year ago
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    so is the answer \[-4-\sqrt{6}\]

  71. anonymous
    • one year ago
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    Somehow your numbers doubled. Should be \[-2-\sqrt3\]

  72. anonymous
    • one year ago
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    so when it's two radicals with the same number inside , it stays the same number ?

  73. anonymous
    • one year ago
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    |dw:1440273681350:dw|

  74. anonymous
    • one year ago
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    |dw:1440273726704:dw|

  75. anonymous
    • one year ago
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    oh okay , I see what I did wrong

  76. anonymous
    • one year ago
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    thank you again

  77. anonymous
    • one year ago
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    you're welcome

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