## anonymous 4 years ago 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??

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

right

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. 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??

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

right