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anonymous

  • 5 years ago

how do I find sin 5ᴨ/8, cos 5ᴨ/8 and tan 5ᴨ/8 using trig identities

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  1. anonymous
    • 5 years ago
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    What is ᴨ

  2. anonymous
    • 5 years ago
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    use formula for sin4x

  3. anonymous
    • 5 years ago
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    the problem doesn't say...just says to use trigonometric identities to find the values

  4. anonymous
    • 5 years ago
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    But what is "ᴨ" is it theta or something else. What is it?

  5. anonymous
    • 5 years ago
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    expand sin4x, put x as 5pi/8 and equate the expa nsion of sin4x to 1

  6. anonymous
    • 5 years ago
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    Is it pi?

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    Sin ((270-45)/2)=sin (5(pi)/8)

  9. anonymous
    • 5 years ago
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    Do you understand this step?

  10. anonymous
    • 5 years ago
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    Do you understand the step "Sin ((270-45)/2)=sin (5(pi)/8)"

  11. anonymous
    • 5 years ago
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    not really, but i think you got the 270 from 3pi/2 and the 45 from pi/4....

  12. anonymous
    • 5 years ago
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    Well, I am trying to express the thing in degree format. (Sorry for using jumbled form). By the term I mean 180 degree

  13. anonymous
    • 5 years ago
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    By the term pi mean 180 degree

  14. anonymous
    • 5 years ago
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    \[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\]

  15. anonymous
    • 5 years ago
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    Now does that makes sense?

  16. anonymous
    • 5 years ago
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    yes

  17. anonymous
    • 5 years ago
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    *make

  18. anonymous
    • 5 years ago
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    Now can you do the rest, or I will have to show you every steps.

  19. anonymous
    • 5 years ago
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    so, sin (225/2)=sin (112.5)? you might have to show me the steps.

  20. anonymous
    • 5 years ago
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    or is it sin (5pi/4 divided by 2)

  21. anonymous
    • 5 years ago
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    Yes, thats true, but thats not helpful here. \[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\] \[=>sin (5\pi/8)=\sqrt(0.5(1-cos(3\pi/2-\pi/4))\]

  22. anonymous
    • 5 years ago
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    Now you got to use addition formula for cos to solve the part with cos, and then it will be done

  23. anonymous
    • 5 years ago
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    I am totally lost .....this does not look at all familiar.....

  24. anonymous
    • 5 years ago
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    are you familiar with double angle formula for trig?

  25. anonymous
    • 5 years ago
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    i'm sure we went over them....but nothing coming to mind

  26. anonymous
    • 5 years ago
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    Then you can't expect to get this thing within your understandability range

  27. anonymous
    • 5 years ago
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    that's why I am having trouble....thought if someone walked me through it, it would make sense

  28. anonymous
    • 5 years ago
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    If you wish I can tutor you (in exchange of some nominal payment) :)

  29. anonymous
    • 5 years ago
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    Seems like you are not interested. So may I leave?

  30. anonymous
    • 5 years ago
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    didn't know i had to pay for help

  31. anonymous
    • 5 years ago
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    No no not that

  32. anonymous
    • 5 years ago
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    you can get free help here for ever

  33. anonymous
    • 5 years ago
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    But I don't think they will be very helpful

  34. anonymous
    • 5 years ago
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    So incase you need personal help

  35. anonymous
    • 5 years ago
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    I mean through audio chatting and virtual black board

  36. anonymous
    • 5 years ago
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    I can't teach you everything here. This place is too inconvenient to type even a simple equation

  37. anonymous
    • 5 years ago
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    I mean no one can

  38. anonymous
    • 5 years ago
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    thanks for the offer, but rather just stay here

  39. anonymous
    • 5 years ago
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    Thats great. No problem

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spraguer (Moderator)
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