Question

Lowering Powers. cos^2xsin^4x

Answer 1

So far I have this:\[(\cos ^{2}x \sin ^{2}x)\sin ^{2}x\]\[(\frac{ 1+\cos 2x }{ 2 })(\frac{ 1-\cos 2x }{ 2 })\times(\frac{ 1-\cos 2x }{ 2 })\]

Answer 2

\[(\frac{ 1-\cos ^{2}2x }{ 4 })\times(\frac{ 1-\cos2x }{ 2 })\]

Answer 3

\[(\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }\cos ^{2}2x)\times(\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }\cos 2x)\]

Answer 4

how did you convert \[\sin ^2 x \] into \[(1-\cos 2x)\div2\]

Answer 5

nm, i see what you did

Answer 6

Looks good to me, now you need to expand reduce cos^22x and expand the double angle of cos2x

Answer 7

I know, but how do I expand cos^22x?

Answer 8

using the same power reducing formula, but you have to include 2x into it.

Answer 9

let your 2x = b if you get confused. Then replace b with 2x when your done.

Answer 10

alright, I'll try it.

Answer 11

So cos^(2)2x = (1+cos4x/2)?

Answer 12

yep

Answer 13

Awesome, thank you!

Answer 14

Do you want an example? I found one from MyMathLab

Answer 15

Sure!

Answer 16

It will take me a few mins to do screen shots and crop the data. Give me like 4mins

Answer 17

Ok

Answer 18

Alright, its going to be 3 pictures from my Mathlab I completed a last month.

Answer 19

Damnit, when i attach a file the post button falls behind the green tab

Answer 20

made it back

Answer 21

There are the other files, i got disconnected from openstudy.

Answer 22

Awesome, thanks again.