Question

Rewrite sin^4x so that it involves only the first powers of cosine.

Answer 1

I have a "calculation" section and a "reason" section.

So far, I have:
\[\frac{ 1 }{ 2 }(1 - \cos(2x)) (\frac{ 1 }{ 2 }(1-\cos2x))\]
Using one of those trig identities... now what? do I distribute?

Answer 2

What do they imply by first powers of cosine? Like 1+x+x^2.....?

Answer 3

Distribute, then apply the same identity to the remaining cos^2 2x

Answer 4

@FaiqRaees they mean no exponents other than 1. Like in terms of cos (2x) and cos (4x) etc

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @agent0smith
@FaiqRaees they mean no exponents other than 1. Like in terms of cos (2x) and cos (4x) etc
\(\color{#0cbb34}{\text{End of Quote}}\)
Oh thanks for clarifying it.

Answer 6

\[(\frac{ 1 }{ 2 } - \frac{ \cos2x }{ 2 })(\frac{ 1 }{ 2 }-\frac{ \cos2x }{ 2 })\]
What remaining cos^2(2x) do you mean? Did you mean the cos2x?

Answer 7

Now multiply the terms together

Answer 8

Upon multiplying the cosine function, you will get a cosine function raised to the power 2. He was talking about that function

Answer 9

Yes. But Abbles, make it easier, not harder
\[\large \frac{ 1 }{ 2 }(1 - \cos(2x)) (\frac{ 1 }{ 2 }(1-\cos2x)) = \frac{ 1 }{ 4 }(1 - \cos(2x))(1-\cos(2x))\]

Answer 10

\[\frac{ 1 }{ 4 }(1-2\cos(2x) + \cos^2(2x))\]
Look about right so far?

Answer 11

Yep, now expand the cos^2 term using the same process as earlier.

Answer 12

\[\frac{ 1 }{ 4 }(1 - 2cosx(2x) + \frac{ 1 + \cos4 }{ 2 })\]
Oh boy...

Answer 13

Next multiply the 1/4 through?

Answer 14

\[\frac{ 1 - 2cosx(2x) + \frac{ 1+\cos4 }{ 2 }}{ 4 }\]

Answer 15

cos 4x but yes. I'd prob just keep the 1/4 outside for neatness.

Answer 16

And you have an x(2x)

Answer 17

cos 4x ?
hah! didn't see the extra x :P

Answer 18

\[\frac{ 1 - 2\cos(2x) }{ 4 } + \frac{ 1 + \cos4 }{ }\]
Could I rewrite it like this?

Answer 19

cos 4x not cos 4 ;P \[\large \frac{ 3 }{ 8 }-\frac{1 }{ 2 }\cos 2x+\frac{ 1 }{ 8 }\cos 4x\]

Answer 20

\[\frac{ 3 - 4\cos(2x) + \cos4x }{ 8 }\]
Is this right?

Answer 21

Yes, same as what i have above.

Answer 22

Okay, thanks - I think this is what they're looking for. For some reason the less doesn't factor out 1/8... is that preferred?

Answer 23

It doesn't matter, depends on the book I guess. Yours or mine look fine.

Answer 24

http://www.wolframalpha.com/input/?i=sin%5E4+x

wolfram gives it with the 1/8 factored but i doubt it matters.

Answer 25

Thanks :)

Answer 26

Welcome :)