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anonymous

  • one year ago

Need help with some trig problems

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @jim_thompson5910

  3. jim_thompson5910
    • one year ago
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    Use the identity cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)

  4. jim_thompson5910
    • one year ago
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    you're given tan(B). Use that to find the value of sin(B)

  5. anonymous
    • one year ago
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    wait we are solving question number 1 right

  6. jim_thompson5910
    • one year ago
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    yes

  7. anonymous
    • one year ago
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    ok how do I do that

  8. jim_thompson5910
    • one year ago
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    draw a right triangle with the angle beta |dw:1438649540480:dw|

  9. jim_thompson5910
    • one year ago
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    if tan(B) = -1/2, then what does that mean for the right triangle above?

  10. anonymous
    • one year ago
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    ok one sec

  11. anonymous
    • one year ago
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    sin is \[\sqrt{5}/5\]

  12. jim_thompson5910
    • one year ago
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    correct |dw:1438649735884:dw|

  13. jim_thompson5910
    • one year ago
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    what's cos(B) equal to?

  14. anonymous
    • one year ago
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    \[\frac{ 2\sqrt{5} }{ 5 }\]

  15. jim_thompson5910
    • one year ago
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    yes

  16. jim_thompson5910
    • one year ago
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    actually, B is in Q2, so cosine is negative

  17. jim_thompson5910
    • one year ago
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    now you do the same for the angle alpha |dw:1438649901995:dw|

  18. jim_thompson5910
    • one year ago
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    |dw:1438649921905:dw|

  19. anonymous
    • one year ago
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    oh yeah

  20. jim_thompson5910
    • one year ago
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    sin(alpha) will be negative since alpha is in Q4

  21. anonymous
    • one year ago
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    \[\sqrt{23}/5\]

  22. jim_thompson5910
    • one year ago
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    that will be negative, but otherwise, good

  23. anonymous
    • one year ago
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    ok

  24. jim_thompson5910
    • one year ago
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    \[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) =??\]

  25. anonymous
    • one year ago
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    so \[\frac{ \sqrt{2} }{ 5 }\times \frac{ -2\sqrt{5} }{ 5 }+-\frac{ \sqrt{23} }{ 5 }\times- \frac{\sqrt{5} }{ 5 }\]

  26. jim_thompson5910
    • one year ago
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    sin(B) is positive because B is in Q2

  27. anonymous
    • one year ago
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    I cant get an exact answer on my calculator

  28. jim_thompson5910
    • one year ago
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    \[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) = \frac{-2\sqrt{2*5}}{5*5} + \frac{-\sqrt{23*5}}{5*5}\] \[\Large \cos(\alpha - \beta) = -\frac{2\sqrt{10}}{25} - \frac{\sqrt{115}}{25}\]

  29. jim_thompson5910
    • one year ago
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    Optionally, you can factor out the GCF \(\Large \frac{\sqrt{5}}{25}\) to get \[\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\] \[\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)\] \[\Large \cos(\alpha - \beta) = \frac{\sqrt{5}}{25}\left(-2\sqrt{2}-\sqrt{23}\right)\]

  30. anonymous
    • one year ago
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    \[\frac{ -2\sqrt{10}-\sqrt{115} }{ 25 }\]

  31. anonymous
    • one year ago
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    is that the answer

  32. jim_thompson5910
    • one year ago
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    yeah that's another way to write the answer

  33. anonymous
    • one year ago
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    ok how do we do the second one

  34. jim_thompson5910
    • one year ago
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    review this list of identities http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look for the "tan(2theta)" identity on the second page

  35. anonymous
    • one year ago
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    ok i see it

  36. jim_thompson5910
    • one year ago
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    so you need to find tan(alpha) and plug it into that identity

  37. anonymous
    • one year ago
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    ok

  38. anonymous
    • one year ago
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    tan is \[\frac{ - \sqrt{46}}{ 2}\]

  39. jim_thompson5910
    • one year ago
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    good

  40. jim_thompson5910
    • one year ago
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    alpha is in Q4 where tan is negative

  41. anonymous
    • one year ago
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    ok so can you show me how you plug it in

  42. jim_thompson5910
    • one year ago
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    \[\Large \tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}\] \[\Large \tan(2\alpha) = \frac{2\left(\frac{-\sqrt{46}}{2}\right)}{1-\left(\frac{-\sqrt{46}}{2}\right)^2}\] \[\Large \tan(2\alpha) = \frac{-\sqrt{46}}{1-\frac{46}{4}}\] I'll let you simplify

  43. anonymous
    • one year ago
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    ok i got\[\frac{ 2\sqrt{46} }{ 21 }\]

  44. jim_thompson5910
    • one year ago
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    me too

  45. anonymous
    • one year ago
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    great does the attachement you sent me have the equation for the third one as well

  46. jim_thompson5910
    • one year ago
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    yes, look under "Sum and Difference Formulas" on page 2

  47. anonymous
    • one year ago
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    sin(a)cos(b)+cos(a)sin(b)

  48. jim_thompson5910
    • one year ago
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    yes

  49. anonymous
    • one year ago
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    ok again I am having problems plugging the equations in here from my side, can you show me what you got when you plugged each of the values in

  50. jim_thompson5910
    • one year ago
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    The values of sin(alpha), sin(beta), cos(alpha), cos(beta) were determined previously. Plug them in to get \[\Large \sin(\beta + \alpha) = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)\] \[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) =??\]

  51. anonymous
    • one year ago
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    I am having a hard time typing it in

  52. anonymous
    • one year ago
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    one second

  53. anonymous
    • one year ago
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    \[-\frac{ -2\sqrt{23}+\sqrt{2}}{ 5\sqrt{5} }\]

  54. jim_thompson5910
    • one year ago
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    \[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{-2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{-2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}-2\sqrt{115}}{25}\]

  55. anonymous
    • one year ago
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    ok I see what I did wrong, I messed up on the second step

  56. jim_thompson5910
    • one year ago
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    oops I should have a positive sign

  57. anonymous
    • one year ago
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    where in front of the 2

  58. jim_thompson5910
    • one year ago
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    \[\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{2\sqrt{5*23}}{5*5}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{2\sqrt{115}}{25}\] \[\Large \sin(\beta + \alpha) = \frac{\sqrt{10}+2\sqrt{115}}{25}\]

  59. jim_thompson5910
    • one year ago
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    yeah

  60. anonymous
    • one year ago
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    ok

  61. anonymous
    • one year ago
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    @jim_thompson5910 i have to go to the bathroom I will be back in about 6 minutes

  62. jim_thompson5910
    • one year ago
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    ok

  63. anonymous
    • one year ago
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    In the meanwhile could you input the information for #5 so I can see how you did it

  64. jim_thompson5910
    • one year ago
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    you mean #4 ?

  65. anonymous
    • one year ago
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    @jim_thompson5910 sorry that took longer than expected

  66. anonymous
    • one year ago
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    ok so I tried finding the equation for question #4 and #5 in the attachment you sent, but I could not find them

  67. jim_thompson5910
    • one year ago
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    alpha is in Q4 270 < alpha < 360 270/2 < alpha/2 < 360/2 135 < alpha/2 < 180 so alpha/2 is in Q2 which means sin(alpha/2) will be positive ------------------------------------------------------- Look on that identity sheet on page 2. Look under the "Half Angle Formulas" section You will use the identity \[\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos(\alpha)}{2}}\] the right side is positive because sin(alpha/2) is positive

  68. anonymous
    • one year ago
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    ok i see it now

  69. anonymous
    • one year ago
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    i just plug in and solve, right?

  70. jim_thompson5910
    • one year ago
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    yes, \[\Large \cos(\alpha) = \frac{\sqrt{2}}{5}\]

  71. anonymous
    • one year ago
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    ok so my answer is

  72. jim_thompson5910
    • one year ago
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    I'm getting \[\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{5-\sqrt{2}}{10}}\] which is the same thing more or less

  73. anonymous
    • one year ago
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    ok sweet, alright last problem

  74. anonymous
    • one year ago
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    its the same thing, but instead of subtracting root, we add this time

  75. anonymous
    • one year ago
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    oh no, we need a different value

  76. jim_thompson5910
    • one year ago
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    you'll use a different identity

  77. jim_thompson5910
    • one year ago
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    and you're using beta instead of alpha

  78. anonymous
    • one year ago
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    yeah ok so I found cosine in terms of beta, which is \[-\frac{ 2\sqrt{5} }{ 5 }\]

  79. jim_thompson5910
    • one year ago
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    yes

  80. anonymous
    • one year ago
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    Is this my equation

  81. jim_thompson5910
    • one year ago
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    now simplify that

  82. anonymous
    • one year ago
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  83. jim_thompson5910
    • one year ago
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    I would rationalize the denominator

  84. jim_thompson5910
    • one year ago
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    multiply top and bottom of that inner fraction by sqrt(5)

  85. jim_thompson5910
    • one year ago
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    doing that gives \[\Large \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{5-2\sqrt{5}}{10}}\]

  86. anonymous
    • one year ago
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    Is that as simplified as it gets?

  87. jim_thompson5910
    • one year ago
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    yeah

  88. anonymous
    • one year ago
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    wait I think one more problem got cut out while I was uploading my attachment

  89. anonymous
    • one year ago
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    Here one sec this is simpler, but it got me confused

  90. jim_thompson5910
    • one year ago
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    make a new post. This one is getting very laggy

  91. anonymous
    • one year ago
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    ok

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