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anonymous

  • 4 years ago

evaluate the six trigonometric functions of the angle (see below for picture)

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    the point is (4, -3) see the picture above

  3. anonymous
    • 4 years ago
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    im just confused because the arrow goes all the way around so i dont know how to find answers

  4. myininaya
    • 4 years ago
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    |dw:1328392741044:dw|

  5. anonymous
    • 4 years ago
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    is that all i have to do

  6. myininaya
    • 4 years ago
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    \[\sin(\theta)=\frac{opp}{hyp}=\frac{-3}{5}\]

  7. anonymous
    • 4 years ago
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    and do i list negatives in the answers

  8. anonymous
    • 4 years ago
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    oh wait

  9. myininaya
    • 4 years ago
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    adj=4 opp=-3 hyp=5

  10. anonymous
    • 4 years ago
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    i forgot thats not the angle. here is a new picture

  11. anonymous
    • 4 years ago
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  12. anonymous
    • 4 years ago
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    feta is on the outside its the big one going around

  13. anonymous
    • 4 years ago
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    ??????

  14. myininaya
    • 4 years ago
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    I'm confused Do you want to find theta or do you want us to evaluate theta using the 6 trig ratios?

  15. anonymous
    • 4 years ago
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    Do the operation myinaya suggested, then add 360 to get the angle that is there

  16. anonymous
    • 4 years ago
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    evalute using ratios. but im confused because the theta is on the outside and not the inside

  17. myininaya
    • 4 years ago
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    let me look at your pic again one sec

  18. anonymous
    • 4 years ago
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    ok look at the second one

  19. myininaya
    • 4 years ago
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    ok

  20. myininaya
    • 4 years ago
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    |dw:1328393266943:dw|

  21. myininaya
    • 4 years ago
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    \[\sin(360^o-\theta^o)=\frac{opp}{hyp}=\frac{-3}{5}\] \[\sin(360)\cos(\theta)-\sin(\theta) \cos(360)=\frac{-3}{5}\]

  22. myininaya
    • 4 years ago
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    \[(0) \cos(\theta)-\sin(\theta)(1)=\frac{-3}{5}\]

  23. myininaya
    • 4 years ago
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    \[\sin(\theta)=\frac{3}{5}\]

  24. myininaya
    • 4 years ago
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    Does that make since?

  25. myininaya
    • 4 years ago
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    sense*?

  26. anonymous
    • 4 years ago
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    no sry i dont get what u just said

  27. myininaya
    • 4 years ago
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    which part?

  28. myininaya
    • 4 years ago
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    There are 360 degrees in circle we have theta already making up part of that 360 degrees 360-theta is left to make a full rotation

  29. myininaya
    • 4 years ago
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    Recall your sum and difference formulas

  30. anonymous
    • 4 years ago
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    ok but how do i solve for sin cos tan etc if its on the outside

  31. myininaya
    • 4 years ago
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    you will need those here

  32. myininaya
    • 4 years ago
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    I just did sin(theta) for you

  33. myininaya
    • 4 years ago
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    \[\cos(360-\theta)=\frac{4}{5}\] \[\cos(360)\cos(\theta)+\sin(360 )\sin(\theta)=\frac{4}{5}\]

  34. myininaya
    • 4 years ago
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    \[1\cos(\theta)+0 \sin(\theta)=\frac{4}{5} => \cos(\theta)=\frac{4}{5}\]

  35. anonymous
    • 4 years ago
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    so is it the same as being on teh inside

  36. myininaya
    • 4 years ago
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    well sin(theta) does not equal sin(360-theta) but cos(theta) equals cos(360-theta) so it is not the same

  37. myininaya
    • 4 years ago
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    it is not the same because it is not always true and i showed this by showing sin(theta) does not equal sin(360-theta)

  38. anonymous
    • 4 years ago
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    but when i do it using theta on teh inside i get the same answers (sin is -3/5, cos is 4/5, tan is -3/4, etc

  39. myininaya
    • 4 years ago
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    no sin(theta)=3/5 sin(360-theta)=-3/5 3/5 does not equal -3/5

  40. anonymous
    • 4 years ago
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    ohh ok....thanks!!!

  41. anonymous
    • 4 years ago
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    can i ask you another question

  42. anonymous
    • 4 years ago
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    its kinda silly...but how do u simplify down (-4/\[\sqrt{52}\])

  43. myininaya
    • 4 years ago
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    what is 13*4

  44. anonymous
    • 4 years ago
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  45. myininaya
    • 4 years ago
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    \[\sqrt{52}=\sqrt{4} \sqrt{13}=2 \sqrt{13}\]

  46. anonymous
    • 4 years ago
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    see pic above...i got it down to there but then stuck

  47. myininaya
    • 4 years ago
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    \[\frac{-4}{2 \sqrt{13}}=\frac{-2}{\sqrt{13}}=\frac{-2 \sqrt{13}}{13}\]

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