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# P2 help needed watch

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?
2. (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?
3. 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
4. (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
5. (Original post by IntegralAnomaly)
well sin2A=2sinAcosA so cosAsin2A=2cos²AsinA
ok.. i got it

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