anonymous
  • anonymous
hello can sumone help me
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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idily101
  • idily101
hi
anonymous
  • anonymous
what do you need help with?
anonymous
  • anonymous
half angle formulas

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anonymous
  • anonymous
in particular a problem my teacher gave me that quite anumber of ppl hav had difficulty wit
anonymous
  • anonymous
I don't think I can help you with that though :( sorry
anonymous
  • anonymous
oh well appreciate ur offer
anonymous
  • anonymous
what kind of half angle formulas?
anonymous
  • anonymous
cos^6(x) i have to rewrite it in terms of the first power of the cosine
anonymous
  • anonymous
did u mean cos(x)^6
anonymous
  • anonymous
no cos^6(x)
anonymous
  • anonymous
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
anonymous
  • anonymous
wate how u get the first part
anonymous
  • anonymous
which first part?
anonymous
  • anonymous
cos(3x) = 4cos^3(x) -3 cos(x), so cos^3(x) = cos(3x) + 3cos(x).
anonymous
  • anonymous
oops...sorry, forgot to divide by 4
anonymous
  • anonymous
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
anonymous
  • anonymous
oh ok thanx
anonymous
  • anonymous
you are welcome

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