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drasy22

  • one year ago

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

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  1. drasy22
    • one year ago
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    \[\cos ^{3}\]

  2. drasy22
    • one year ago
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    cos^3theta

  3. drasy22
    • one year ago
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    \[\cos ^{3}\theta \]

  4. campbell_st
    • one year ago
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    it is simply \[\cos(\theta)\cos(\theta)\cos(\theta)\]

  5. drasy22
    • one year ago
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    no i have to use like identities

  6. drasy22
    • one year ago
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    ex if it is sin ^4 u do sin2)^2 and then you put the identitiy for sin^2 and then solve

  7. drasy22
    • one year ago
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    but i have no idea how to do this

  8. drasy22
    • one year ago
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    @zepdrix

  9. zepdrix
    • one year ago
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    Hmm I'm not familiar with any identities that decrease the power on trig functions. Are you sure it wasn't suppose to be \(cos (3\theta)\) ?

  10. drasy22
    • one year ago
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    yes i am sure but thanks

  11. campbell_st
    • one year ago
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    well start with \[\cos(\theta) (\cos^2(\theta)) = \cos(\theta)(1 - \sin^2(\theta))\]

  12. drasy22
    • one year ago
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    hmmm k

  13. zepdrix
    • one year ago
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    Oh I guess we could use the Half-Angle Formulas to decrease power. I somehow forgot about those when I made my last comment :) lol

  14. zepdrix
    • one year ago
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    Here's a helpful identity we can use. \[\large \color{royalblue}{\cos^2x=\frac{1+\cos2x}{2}}\]

  15. drasy22
    • one year ago
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    yeah u are right i will have to use the power reducing identities

  16. drasy22
    • one year ago
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    yeah i started by using that but i don't know what to do after that step

  17. zepdrix
    • one year ago
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    \[\large \cos^3x\quad =\quad \cos x \color{royalblue}{\cos^2 x} \quad = \quad \cos x \color{royalblue}{\frac{1+\cos2x}{2}}\]

  18. zepdrix
    • one year ago
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    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.

  19. drasy22
    • one year ago
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    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!!!

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