anonymous
  • anonymous
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??
Mathematics
chestercat
  • chestercat
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dumbcow
  • dumbcow
yes i believe you do have to use binomial expansion \[\cos(6 \theta)+\sin(6 \theta) = (\cos \theta +\sin \theta)^{6}\]
anonymous
  • anonymous
but it says using demoivre..so binomial isnt needed i think'
anonymous
  • anonymous
isn't needed?? i don't understand..

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dumbcow
  • dumbcow
wait i forgot the "i" the equality only holds if dealing with complex numbers \[\cos(6 \theta)+i \sin(6 \theta) = (\cos \theta +i \sin \theta)^{6}\]
anonymous
  • anonymous
right
anonymous
  • anonymous
i always did like what @dumbcow did.. is there other alternative??
anonymous
  • anonymous
if it says without demoivre,,then there are other wayss
dumbcow
  • dumbcow
\[=\cos^{6}+6i \cos^{5}\sin-15\cos^{4}\sin^{2}-20i \cos^{3}\sin^{3}+15\cos^{2}\sin^{4}+6i \cos \sin^{5}-\sin^{6}\]
anonymous
  • anonymous
i think you dont need the expansion,as it said find in terms of powers of costheta
anonymous
  • anonymous
find cos6theta in terms of costheta and sin6theta in terms of sintheta,are these separate or same question?
anonymous
  • anonymous
find cos6θ and sin 6θ in powers of cos θ and sin θ.. the same.. i think i have to use the binomial..
anonymous
  • anonymous
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??
anonymous
  • anonymous
the question is find cos6theta and sin6theta , it indicates they are different sums........ it is "and", why take it as +
dumbcow
  • dumbcow
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
anonymous
  • anonymous
Thanks a lot!! Im not lost anymore.. ^^
anonymous
  • anonymous
right

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