hello can sumone help me

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hello can sumone help me

Mathematics
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hi
what do you need help with?
half angle formulas

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

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