amaths, ...can anyone help me? |
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 Sep 9 2005, 09:53 AM
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#1
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 Senior Member  Group: Member Posts: 64 Joined: Feb 2005 Member No: 104,345  |
i'm calculation (actually proving) an amaths question.
when it nearly comes to the end, i dunno how to do, so i looked at the answer. and my question is, how can [sin2kθ (1-2sin^2(i mean square here)θ ) +sin2θcos2kθ ] / 2sinθ change into [sin(2k+2)θ ] / 2sinθ ? i would be grateful if someone can tell me how this can be converted. thanks edit// well...er....i'm in grade 11(form 5) now. yeah. jus to let u know... |
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| *xcaitlinx* |
Sep 9 2005, 02:12 PM
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#2
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Guest |
woahh..that looks liek complete gibberish to me! well i certainly don't know how to do that...maybe it's because im only a freshmen.
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Sep 9 2005, 10:09 PM
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#3
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 hi. call me linda. Group: Official Member Posts: 8,187 Joined: Feb 2004 Member No: 3,475 |
Ahhh trig and trig identities... I'm working on it now...
I'll edit this when I figure it out. |
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Sep 12 2005, 07:02 AM
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#4
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Senior Member Group: Member Posts: 64 Joined: Feb 2005 Member No: 104,345 |
urgh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
well... anyone? please? |
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Sep 22 2005, 06:38 AM
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#5
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Senior Member Group: Member Posts: 64 Joined: Feb 2005 Member No: 104,345 |
ok, i think i can finally know y now!!!!!!! = =" no one reply me
  (finally, figured this out. so hapi) if you wan to know how, c the followingas cos2θ=(1-2sin^2 θ) so [sin2kθ (1-2sin^2 θ ) +sin2θcos2kθ ] / 2sinθ =[sin2kθcos2θ+sin2θcos2kθ ] / 2sinθ according to the Compound Angle Formula sin(A+B)=sinAcosB+cosAsinB so [sin2kθcos2θ+sin2θcos2kθ ] / 2sinθ = sin(2kθ+2θ)/2sinθ = sin(2k+2)θ/2sinθ = answer 
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| *mzkandi* |
Sep 24 2005, 02:04 PM
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#6
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Ok, this question was asked way back on Sept. 9, so I will close this.
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