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

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Other answers:

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

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