Question

express (costheta + isintheta/sintheta + icostheta)^6 in the form of x+iy.​

Answer 1

Step-by-step explanation:

(costheta + isintheta/sintheta + icostheta)^6

=(costheta + isintheta)^6 / (sintheta + icostheta)^6

=(cos6theta + isin6theta)/ ((cos(6pi/2-6theta)+isin(6pi2-6theta))

=(cos6theta + isin6theta)/(cos((540-6theta)+isin(540-6theta))

=(cos6theta + isin6theta)/(cos((180-6theta)+isin(180-6theta))

=(cos6theta + isin6theta)/(-cos(6theta)+isin(6theta))

=(cos6theta + isin6theta)/(-cos(6theta)+isin(6theta)) * (cos6theta + isin6theta)/(cos(6theta)+isin(6theta)

=(cos6theta + isin6theta)^2/((isin(6theta))^2-(cos6theta)^2)

=(cos12theta + isin12theta)/(-(sint6theta)^2-(cos6theta)^2)

here i^2=- 1

=(cos12theta + isin12theta)/(-((sint6theta)^2+(cos6theta)^2))

=(cos12theta + isin12theta)/-1

=(-cos12theta)+i(-sin12theta)