Question

hello can sumone help me

Answer 1

hi

Answer 2

what do you need help with?

Answer 3

half angle formulas

Answer 4

in particular a problem my teacher gave me that quite anumber of ppl hav had difficulty wit

Answer 5

I don't think I can help you with that though :( sorry

Answer 6

oh well appreciate ur offer

Answer 7

what kind of half angle formulas?

Answer 8

cos^6(x) i have to rewrite it in terms of the first power of the cosine

Answer 9

did u mean cos(x)^6

Answer 10

no cos^6(x)

Answer 11

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

Answer 12

wate how u get the first part

Answer 13

which first part?

Answer 14

cos(3x) = 4cos^3(x) -3 cos(x), so cos^3(x) = cos(3x) + 3cos(x).

Answer 15

oops...sorry, forgot to divide by 4

Answer 16

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

Answer 17

oh ok thanx

Answer 18

you are welcome