anonymous
  • anonymous
sin^2(3x)=1−cos6x , can someone explain the process?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
sin^2(3x) = 1 - cos^2(3x) might be useful here
anonymous
  • anonymous
no start with cos 6x = 1 - 2sin^2(3x) 1 - cos 6x = 2sin^2(3x) - are you sure you have the correct expression
anonymous
  • anonymous
Jimmyrep had the right idea...

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anonymous
  • anonymous
\[1-\cos(6x)=2\sin^23x\]
anonymous
  • anonymous
Oh .
anonymous
  • anonymous
Not so bright is correct...to sub
anonymous
  • anonymous
\[\sin^23x=0\] \[3x=n \pi\]
anonymous
  • anonymous
yes i think notsobright is correct
anonymous
  • anonymous
cos(2u)=1-2sin^2(u)
anonymous
  • anonymous
yup
anonymous
  • anonymous
since u = 3x then cos2(3x)= 1-2sin^2(3x)
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
this is what jimmyrep had to start with...not what is given...make sure you copied it down correctly.
anonymous
  • anonymous
Alright .
anonymous
  • anonymous
notsobright is correct
anonymous
  • anonymous
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...
anonymous
  • anonymous
values in which the sin^2(3x)=0...or when 3x=npi as was stated. sorry for the confusion.
anonymous
  • anonymous
Got it , thanks again !

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