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anonymous

  • 5 years ago

hello can sumone help me

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  1. idily101
    • 5 years ago
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    hi

  2. anonymous
    • 5 years ago
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    what do you need help with?

  3. anonymous
    • 5 years ago
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    half angle formulas

  4. anonymous
    • 5 years ago
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    in particular a problem my teacher gave me that quite anumber of ppl hav had difficulty wit

  5. anonymous
    • 5 years ago
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    I don't think I can help you with that though :( sorry

  6. anonymous
    • 5 years ago
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    oh well appreciate ur offer

  7. anonymous
    • 5 years ago
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    what kind of half angle formulas?

  8. anonymous
    • 5 years ago
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    cos^6(x) i have to rewrite it in terms of the first power of the cosine

  9. anonymous
    • 5 years ago
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    did u mean cos(x)^6

  10. anonymous
    • 5 years ago
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    no cos^6(x)

  11. anonymous
    • 5 years ago
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    We know that cos(3x) = 4cos^3(x) -3 cos(x), so cos^3(x) = cos(3x) + 3cos(x). Thus cos^6(x) =[cos(3x) + 3cos(x)]^2 = cos^2(3x) + 6cos(3x)cos(x) + 9cos^2(x) Then we know cos(2x) = 2cos^2(x) -1, so cos^2(x) = (1/2)[cos(2x) +1]. Also so cos^2(3x) = (1/2)[cos(6x) +1]. I guess you can finish it up

  12. anonymous
    • 5 years ago
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    wate how u get the first part

  13. anonymous
    • 5 years ago
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    which first part?

  14. anonymous
    • 5 years ago
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    cos(3x) = 4cos^3(x) -3 cos(x), so cos^3(x) = cos(3x) + 3cos(x).

  15. anonymous
    • 5 years ago
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    oops...sorry, forgot to divide by 4

  16. anonymous
    • 5 years ago
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    4cos^3(x) = cos(3x) + 3cos(x) So 16cos^6(x) =[cos(3x) + 3cos(x)]^2 you continue the same thing on the righr hand side. After you get everything, then you divide both sides with 16

  17. anonymous
    • 5 years ago
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    oh ok thanx

  18. anonymous
    • 5 years ago
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    you are welcome

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spraguer (Moderator)
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