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anonymous

  • one year ago

Need some Trig HElp

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  1. anonymous
    • one year ago
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    what do you need help with?

  2. anonymous
    • one year ago
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    i will attach the file, give me a second

  3. anonymous
    • one year ago
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  4. anonymous
    • one year ago
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    @dan815

  5. anonymous
    • one year ago
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    @jim_thompson5910

  6. DanJS
    • one year ago
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    hmm, try replacing cos^2 with 1 - sin^2

  7. anonymous
    • one year ago
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    which attachment are you on

  8. DanJS
    • one year ago
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    the first one...

  9. DanJS
    • one year ago
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    \[3\sin(\theta)+3 = 2\cos^2(\theta)\]

  10. anonymous
    • one year ago
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    ok

  11. anonymous
    • one year ago
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    i did that

  12. DanJS
    • one year ago
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    ok, it looks like you can factor that now if you put it in ax^2+bx+c form , with the x being sin

  13. DanJS
    • one year ago
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    \[2\sin^2(\theta)+3\sin(\theta)+1=0\]

  14. anonymous
    • one year ago
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    wait but this is what my equation looks like so far... is it right

  15. anonymous
    • one year ago
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    @DanJS

  16. DanJS
    • one year ago
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    right, from the unit circle , there is the property, sin^2 + cos^2 = 1, that is what we did

  17. DanJS
    • one year ago
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    then just simplify that down to

  18. DanJS
    • one year ago
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    \[2\sin^2(\theta)+3\sin(\theta)+1=0\]

  19. anonymous
    • one year ago
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    ok got it

  20. DanJS
    • one year ago
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    so you can factor that like usual quadratics , let v=sin(theta) 2v^2 + 3v + 1 = 0

  21. anonymous
    • one year ago
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    ok give me a sec

  22. DanJS
    • one year ago
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    \[(v+1)(2v+1)=0\]

  23. DanJS
    • one year ago
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    replace v back with sin(theta) \[(\sin \theta + 1)(2\sin \theta +1) = 0\]

  24. DanJS
    • one year ago
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    good so far?

  25. anonymous
    • one year ago
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    yeah

  26. DanJS
    • one year ago
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    that is true if either of those parenthesis terms are zero, so you have \[\sin(\theta) = -1~~~or~~~\sin(\theta) = -1/2\]

  27. DanJS
    • one year ago
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    do you know how to figure those from the unit circle?

  28. anonymous
    • one year ago
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    yeah, but im kind of weak in my unit circle so i might get it wrong

  29. DanJS
    • one year ago
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    Ill tell ya, soon as you get that picture, everything clicks good...

  30. anonymous
    • one year ago
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    ok should I gust google unit circle

  31. DanJS
    • one year ago
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    http://www.embeddedmath.com/downloads/files/unitcircle/unitcircle-letter.pdf

  32. DanJS
    • one year ago
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    do you want a quick rundown of it, how to determine angles

  33. anonymous
    • one year ago
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    no i got it

  34. anonymous
    • one year ago
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    i just needed a brush up

  35. anonymous
    • one year ago
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    the picture is fine

  36. DanJS
    • one year ago
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    aight, what are the answers for this one

  37. anonymous
    • one year ago
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    the answers are -pi/2, 3pi/2, -pi/6, 7p/6

  38. DanJS
    • one year ago
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    read the problem one more time through...

  39. anonymous
    • one year ago
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    oh so it is just 3pi/2 and 7pi/6

  40. DanJS
    • one year ago
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    not quite... lets look at the first one... sin(theta) = -1 Each point on the circle is the coordinate (cos , sin) , the y coordinate is the sin of the angle... y is -1 at 270 or 3pi/2

  41. DanJS
    • one year ago
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    for sin(theta) = -1/2 there are 2 angle values where the y coordinate is -1/2

  42. DanJS
    • one year ago
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    |dw:1438716059055:dw|

  43. anonymous
    • one year ago
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    ahhhh

  44. DanJS
    • one year ago
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    i remember taking trig and never got how angles were figured till this circle pic , after the class was over

  45. DanJS
    • one year ago
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    radius is 1, so sin = opp/hyp = opp/1 = opp , or the y coordinate

  46. anonymous
    • one year ago
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    ok that makes sense

  47. DanJS
    • one year ago
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  48. anonymous
    • one year ago
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    thats helpful thanks

  49. DanJS
    • one year ago
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    welcome

  50. anonymous
    • one year ago
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    so the solutions to the equation are?

  51. anonymous
    • one year ago
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    @DanJS

  52. anonymous
    • one year ago
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    @Loser66 can you help me with my other questions

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