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anonymous
 5 years ago
hello can sumone help me
anonymous
 5 years ago
hello can sumone help me

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you need help with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in particular a problem my teacher gave me that quite anumber of ppl hav had difficulty wit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think I can help you with that though :( sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well appreciate ur offer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what kind of half angle formulas?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos^6(x) i have to rewrite it in terms of the first power of the cosine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We 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
 5 years ago
Best ResponseYou've already chosen the best response.0wate how u get the first part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(3x) = 4cos^3(x) 3 cos(x), so cos^3(x) = cos(3x) + 3cos(x).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops...sorry, forgot to divide by 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04cos^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
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