## Mandy_Nakamoto 3 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. dumbcow

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

2. Soham051994

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

3. Mandy_Nakamoto

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

4. dumbcow

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

5. Soham051994

right

6. Mandy_Nakamoto

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

7. Soham051994

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

8. dumbcow

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

9. Soham051994

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

10. Soham051994

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

11. Mandy_Nakamoto

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

12. Mandy_Nakamoto

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. Soham051994

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

14. 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

15. Mandy_Nakamoto

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

16. Soham051994

right