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OpenStudy (anonymous):
Use the Double-Angle Formulas to rewrite cos4x in terms of cos x
HELP. I knw the steps but ill explain more in comments...
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OpenStudy (anonymous):
I get to 2(cos2x)^2-1...
OpenStudy (anonymous):
@primeralph :/ please
OpenStudy (primeralph):
This is the kind of thing where you just keep going until you're done. Not hard.
OpenStudy (anonymous):
but i dont see how you do the next step
OpenStudy (anonymous):
the next thing i should get is...2(2cos2x - 1)^2 - 1
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OpenStudy (anonymous):
Recall that
\[\cos(A+B) = \cos(A)\cos(B) -\sin(A)\sin(B)\]
so cos(4x) = cos (2x + 2x) right? now substitute. and solve
OpenStudy (primeralph):
Do the same for cos2x.
OpenStudy (anonymous):
how is my question
OpenStudy (primeralph):
Tell me what you get.
OpenStudy (anonymous):
ok...thinking
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OpenStudy (anonymous):
2(cos^2u-1)-1
OpenStudy (anonymous):
oh parenthhesis squared
OpenStudy (primeralph):
u?
OpenStudy (anonymous):
x
OpenStudy (anonymous):
sorry....
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OpenStudy (primeralph):
Then expand or something.
OpenStudy (anonymous):
so then 2(4cos^4x-4cos^2x+1)-1
OpenStudy (anonymous):
distribute...
OpenStudy (anonymous):
8cos^4x-8cos^2+1?
OpenStudy (anonymous):
that was the answer :D
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OpenStudy (anonymous):
i wish i could figure this out myself...thanks guys
OpenStudy (primeralph):
Hold on.
OpenStudy (anonymous):
ok
OpenStudy (primeralph):
Let me see all the options.
OpenStudy (anonymous):
that was the only option..it wasnt for a test..i dont cheat like that...this was me trying to understand an example..
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OpenStudy (anonymous):
that was the answer provided
OpenStudy (primeralph):
Okay.
OpenStudy (anonymous):
thank you :)
OpenStudy (primeralph):
You're welcome.
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