Question

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??

Answer 1

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

Answer 2

but it says using demoivre..so binomial isnt needed i think'

Answer 3

isn't needed?? i don't understand..

Answer 4

wait i forgot the "i"
the equality only holds if dealing with complex numbers
\[\cos(6 \theta)+i \sin(6 \theta) = (\cos \theta +i \sin \theta)^{6}\]

Answer 5

right

Answer 6

i always did like what @dumbcow did.. is there other alternative??

Answer 7

if it says without demoivre,,then there are other wayss

Answer 8

\[=\cos^{6}+6i \cos^{5}\sin-15\cos^{4}\sin^{2}-20i \cos^{3}\sin^{3}+15\cos^{2}\sin^{4}+6i \cos \sin^{5}-\sin^{6}\]

Answer 9

i think you dont need the expansion,as it said find in terms of powers of costheta

Answer 10

find cos6theta in terms of costheta and sin6theta in terms of sintheta,are these separate or same question?

Answer 11

find cos6θ and sin 6θ in powers of cos θ and sin θ.. the same.. i think i have to use the binomial..

Answer 12

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??

Answer 13

the question is find cos6theta and sin6theta , it indicates they are different sums........ it is "and", why take it as +

Answer 14

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

Answer 15

Thanks a lot!! Im not lost anymore.. ^^

Answer 16

right