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PLZ HELP#VERYCONFUSED rewrite each expression in term with no power greater than 1
 one year ago
 one year ago
PLZ HELP#VERYCONFUSED rewrite each expression in term with no power greater than 1
 one year ago
 one year ago

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campbell_stBest ResponseYou've already chosen the best response.0
it is simply \[\cos(\theta)\cos(\theta)\cos(\theta)\]
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
no i have to use like identities
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
ex if it is sin ^4 u do sin2)^2 and then you put the identitiy for sin^2 and then solve
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
but i have no idea how to do this
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm I'm not familiar with any identities that decrease the power on trig functions. Are you sure it wasn't suppose to be \(cos (3\theta)\) ?
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
yes i am sure but thanks
 one year ago

campbell_stBest ResponseYou've already chosen the best response.0
well start with \[\cos(\theta) (\cos^2(\theta)) = \cos(\theta)(1  \sin^2(\theta))\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Oh I guess we could use the HalfAngle Formulas to decrease power. I somehow forgot about those when I made my last comment :) lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Here's a helpful identity we can use. \[\large \color{royalblue}{\cos^2x=\frac{1+\cos2x}{2}}\]
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
yeah u are right i will have to use the power reducing identities
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
yeah i started by using that but i don't know what to do after that step
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \cos^3x\quad =\quad \cos x \color{royalblue}{\cos^2 x} \quad = \quad \cos x \color{royalblue}{\frac{1+\cos2x}{2}}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Im not sure if that's exactly what you're looking for. But something to notice here is that we no longer even have a squared cosine, since the two cosines we're left with have different INSIDES, we can't combine them that way.
 one year ago

drasy22Best ResponseYou've already chosen the best response.0
yes i am looking for this but i think it is OK i will ask my teacher. but thanks both of u for ur help!!!
 one year ago
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