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anonymous

  • 4 years ago

sin^2(3x)=1−cos6x , can someone explain the process?

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  1. anonymous
    • 4 years ago
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    sin^2(3x) = 1 - cos^2(3x) might be useful here

  2. anonymous
    • 4 years ago
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    no start with cos 6x = 1 - 2sin^2(3x) 1 - cos 6x = 2sin^2(3x) - are you sure you have the correct expression

  3. anonymous
    • 4 years ago
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    Jimmyrep had the right idea...

  4. anonymous
    • 4 years ago
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    \[1-\cos(6x)=2\sin^23x\]

  5. anonymous
    • 4 years ago
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    Oh .

  6. anonymous
    • 4 years ago
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    Not so bright is correct...to sub

  7. anonymous
    • 4 years ago
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    \[\sin^23x=0\] \[3x=n \pi\]

  8. anonymous
    • 4 years ago
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    yes i think notsobright is correct

  9. anonymous
    • 4 years ago
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    cos(2u)=1-2sin^2(u)

  10. anonymous
    • 4 years ago
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    yup

  11. anonymous
    • 4 years ago
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    since u = 3x then cos2(3x)= 1-2sin^2(3x)

  12. anonymous
    • 4 years ago
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    Thanks!

  13. anonymous
    • 4 years ago
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    this is what jimmyrep had to start with...not what is given...make sure you copied it down correctly.

  14. anonymous
    • 4 years ago
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    Alright .

  15. anonymous
    • 4 years ago
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    notsobright is correct

  16. anonymous
    • 4 years ago
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    I first thought he/she were trying to verify that they were equal.... but as notsobright and kingkos has indicated...it appears that you were solving for x in which case not so bright was correct...

  17. anonymous
    • 4 years ago
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    values in which the sin^2(3x)=0...or when 3x=npi as was stated. sorry for the confusion.

  18. anonymous
    • 4 years ago
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    Got it , thanks again !

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spraguer (Moderator)
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