Recent content by clawkz

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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    thank you all who helped or tried to. I finally understand. The problem I was having was thinking 2(2sinxcosx)-->(4sin2xcos2x), but now i understand it just becomes 4sinxcosx. Really appreciate the help
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    ok makes more sense now, but now that i have 2(2sinxcosx)(cos2x)...does the 2 way in front get distrubuted like this--> (4sinxcosx)(cos2x) why not like this (4sin2xcos2x)(cos2x)?
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    ok well here it goes... sin4x=(4sinxcosx)(1-2sin^2x)-->sin4x--> sin2(2x)--> 2(2sinxcosx)-->4sin2xcos2x-->...the confusing part-->4(2sinxcosx)(1-sin^2x)-->(4sinxcosx)(1-sin^2x) on the italic part am i supposed to distribute the 2 out to 2sinxcosx so i can get the 4sin2xcos2x so that i can...
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    ok I think I understand it now. Can I ask you one last favor and please write it down as if you had just gotten the question on a quiz or something? I want to see every step by step. so 1) Prove : sin4x=(4sinxcosx)(1-2sin^2x) please Its late where I live and I need some sleep.
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    I did that because isn't sin4x the same thing as sin2x + sin 2x? and according to my double angle formula sheet sin2x= 2sinxcosx, so this is why i did 2sinxcosx +2sinxcosx = 4sin2xcos2x . I don't really understand what you mean by "sin(2x) in terms of x" I am not the greatest in math -.- .
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    I thought I had been right but I was still lost. can you please look over my work and tell me where I went wrong. thanks sin4x=(4sinxcosx)(1-sin^2x)--->sin4x= sin(2x+2x)--->2sinxcosx+2sinxcosx= 4sin2xcos2x-->4(2sinxcosx)(cos2x)----> I don't know how 4(2sinxcosx) is equal to (4sinxcosx).
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    never mind i get it. I was over thinking it lol
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    ok i get this-->4sin2xcos2x-->4(2sinxcosx)(1-2sin^2x)--> was I supposed to do something to the bolded 4? as the final thing is (4sinxcosx)(1-2sin^2x)
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    so it would be 2sinxcosx+2sinxcosx--->4sin2xcos2x-->is this right so far? i think I am wrong on the last part...btw I am horrible at precalc
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    Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

    Homework Statement sin4x=(4sinxcosx)(1-2sin^2x) Homework Equations Trig identities. The Attempt at a Solution sin4x=(4sinxcosx)(1-2sin^2x) (4sinxcosx)(cos2x) stuck right here...
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