Question

helpss

Answer 1

Rewrite as a sum of terms containing first power of cosines cos^4pi

Answer 2

@sdfgsdfgs Hey xP pweeze help?

Answer 3

cos^2p

Answer 4

factoring out (cos^2p)(cos^2p)

Answer 5

Using identity
cos^2x=(1+cos^2x)/2)

Answer 6

we get \[(\frac{ 1+\cos^2x }{ 2 })(\frac{ 1+\cos^2x }{ 2})\]

Answer 7

does that work? @zepdrix

Answer 8

sorry that's just cos2x

Answer 9

so it would end up
\[(\frac{ 1+\cos2x }{ 2})^2\]

Answer 10

ya that seems like a good path to take! :)

Answer 11

wats the Q? cos^4pi doesn't make sense as cos(pi)=1 so cos^4pi=1 as well.

Answer 12

No no it's just cos^2p not pi :P p is a variable.

Answer 13

Understand this step?
\[\Large\rm \left(\frac{1+\cos2p}{2}\right)^2=\frac{1}{4}(1+\cos2p)^2\]Then you'll have to expand out the square, and apply your Half-Angle Formula a few more times.

Answer 14

Err maybe just one more time.

Answer 15

hm how would that look?

Answer 16

ok then u can reinse n repeat to further reduce the cos^2(2p) into cos(4p)

Answer 17

\[\Large\rm (1+\cos2p)^2=(1+\cos2p)(1+\cos2p)\]Multiply the stuff girl :O

Answer 18

boy, what happened to the one fourth?!

Answer 19

and wont' that re ^2 it?

Answer 20

1/4(1+cos2p+cos2p+cos^2p)