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Mandy_Nakamoto Group Title

Using de Moivre’s theorem, express cos 6θ and sin 6θ in powers of cos θ and sin θ. Do i have to use binomial expansion??

  • 2 years ago
  • 2 years ago

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  1. dumbcow Group Title
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    yes i believe you do have to use binomial expansion \[\cos(6 \theta)+\sin(6 \theta) = (\cos \theta +\sin \theta)^{6}\]

    • 2 years ago
  2. Soham051994 Group Title
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    but it says using demoivre..so binomial isnt needed i think'

    • 2 years ago
  3. Mandy_Nakamoto Group Title
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    isn't needed?? i don't understand..

    • 2 years ago
  4. dumbcow Group Title
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    wait i forgot the "i" the equality only holds if dealing with complex numbers \[\cos(6 \theta)+i \sin(6 \theta) = (\cos \theta +i \sin \theta)^{6}\]

    • 2 years ago
  5. Soham051994 Group Title
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    right

    • 2 years ago
  6. Mandy_Nakamoto Group Title
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    i always did like what @dumbcow did.. is there other alternative??

    • 2 years ago
  7. Soham051994 Group Title
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    if it says without demoivre,,then there are other wayss

    • 2 years ago
  8. dumbcow Group Title
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    \[=\cos^{6}+6i \cos^{5}\sin-15\cos^{4}\sin^{2}-20i \cos^{3}\sin^{3}+15\cos^{2}\sin^{4}+6i \cos \sin^{5}-\sin^{6}\]

    • 2 years ago
  9. Soham051994 Group Title
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    i think you dont need the expansion,as it said find in terms of powers of costheta

    • 2 years ago
  10. Soham051994 Group Title
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    find cos6theta in terms of costheta and sin6theta in terms of sintheta,are these separate or same question?

    • 2 years ago
  11. Mandy_Nakamoto Group Title
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    find cos6θ and sin 6θ in powers of cos θ and sin θ.. the same.. i think i have to use the binomial..

    • 2 years ago
  12. Mandy_Nakamoto Group Title
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    so i just have to expand it and that's it?? Do i have to separate it between the real and the imaginary in the end??

    • 2 years ago
  13. Soham051994 Group Title
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    the question is find cos6theta and sin6theta , it indicates they are different sums........ it is "and", why take it as +

    • 2 years ago
  14. dumbcow Group Title
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    soham, because de'moivres theorem is mentioned i think it was implied mandy, yes i believe you would have to separate real and imaginary coefficients

    • 2 years ago
  15. Mandy_Nakamoto Group Title
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    Thanks a lot!! Im not lost anymore.. ^^

    • 2 years ago
  16. Soham051994 Group Title
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    right

    • 2 years ago
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