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New Member
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Jul 13, 2012, 12:24 AM
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(sin2x)^2/1-cos2x=2 (cosx)^2
proof that left side = right side
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Uber Member
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Jul 13, 2012, 02:31 AM
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First thing to notice is that the left side has double angles whereas the right hasn't. So, you can start by breaking the right side down to single angles.
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New Member
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Jul 13, 2012, 12:40 PM
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Well I try and I couldn't solve it
I try both sides and I couldn't get the right answer
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Uber Member
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Jul 13, 2012, 12:41 PM
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Could you post what you tried?
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New Member
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Jul 13, 2012, 12:51 PM
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(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-2(sinx)^2)
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New Member
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Jul 13, 2012, 12:53 PM
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I try this side and on the half way I stopped
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Uber Member
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Jul 13, 2012, 12:56 PM
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Originally Posted by sally aa
(sin2x)^2/1-cos2x=(2sinxcosx)(2sinxcosx)/1-(1-(sinx)^2)
Remember that:
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New Member
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Jul 13, 2012, 01:00 PM
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2(sinx)^2 =2sin^2(x) yes I know
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Uber Member
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Jul 13, 2012, 01:01 PM
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Okay, now expand the denominator. What do you get?
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New Member
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Jul 13, 2012, 01:06 PM
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(2sinxcosx)(2sinxcosx)/1-1+2sin^2(x)
=(2sinxcosx)(2sinxcosx)/2sin^2(x)
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Uber Member
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Jul 13, 2012, 01:07 PM
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Good. Now what can you cross out?
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New Member
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Jul 13, 2012, 01:11 PM
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(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?
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New Member
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Jul 13, 2012, 01:14 PM
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(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) that's what I get
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Uber Member
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Jul 13, 2012, 01:14 PM
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Originally Posted by sally aa
(2sinxcosx)(2sinxcosx)=2sin^2(x) the 2sin^2(x) with 2sin(x) ?
I don't understand what you mean here...
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Uber Member
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Jul 13, 2012, 01:15 PM
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Originally Posted by sally aa
(4sinxcosx)/2sin^2(x) =2cos(x)/sin(x) thats what I get
The bolded part is wrong.
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New Member
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Jul 13, 2012, 01:19 PM
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What do u mean ?
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Uber Member
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Jul 13, 2012, 01:20 PM
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I mean:
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New Member
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Jul 13, 2012, 01:23 PM
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sorry its 4sin^2(x)cos^2(x)
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Uber Member
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Jul 13, 2012, 01:24 PM
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Okay, so now you have:
What can you simplify?
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New Member
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Jul 13, 2012, 01:26 PM
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4sin^2(x)cos^2(x)/2sin^2(x) then if I canceled the 4sin^2(x) with 2 sin^2(x) we end up with 2 cos^2(x) is this right ?
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