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etomac
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#1
Report Thread starter 15 years ago
#1
1, obtain an expression for sin3A in terms of sin A
2, obtain an expression for cos3A in terms of cos A
pg 70-71 (Heinemann Pure Mathematics 2 Ex 4C)

how do I do that?
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IntegralAnomaly
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#2
Report 15 years ago
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(Original post by etomac)
1, obtain an expression for sin3A in terms of sin A
2, obtain an expression for cos3A in terms of cos A
pg 70-71 (Heinemann Pure Mathematics 2 Ex 4C)

how do I do that?
use sin3A=sin(A+2A) and cos3A=cos(A+2A).Do u know what to do next?
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etomac
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#3
Report Thread starter 15 years ago
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yes thats what i did
for sin
sinAcos2A + cosAsin2A
sinA(1-2sin^2A) + sin2AcosA
but i dunno how to get rid of cosA and end up with a cubic function
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IntegralAnomaly
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Report 15 years ago
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(Original post by etomac)
yes thats what i did
for sin
sinAcos2A + cosAsin2A
sinA(1-2sin^2A) + sin2AcosA
but i dunno how to get rid of cosA and end up with a cubic function
well sin2A=2sinAcosA so cosAsin2A=2cos²AsinA
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etomac
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#5
Report Thread starter 15 years ago
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(Original post by IntegralAnomaly)
well sin2A=2sinAcosA so cosAsin2A=2cos²AsinA
ok.. i got it
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