Express the following in `a+i b` form: `((cos2theta-isin2theta)^4(cos4theta+isin4theta)^(-5))/((cos3theta+isin3theta)^(-2)(cos3theta-isin3theta)^(-9))`

(cos3theta+isin3theta)^5/(costheta+isintheta)^6

((cos2 theta+i sin2 theta)^(3)(cos3 theta-i sin3 theta)^(5))/((cos3 theta+i sin3 theta)^(2)(cos4 theta+i sin4 theta)^(3))=

Show that, ((cos2theta+isin2theta)^3(cos3theta-isin3theta)^4)/((cos3theta+isin3theta)^2(cos4theta+isin4theta)^(-3))=1

Express the following in a + ib form: (i) ((cos theta+ isin theta)/(sin theta + icos theta))^(4) (ii) ((cos 2theta -isin2theta)^(4)(cos4theta +isin4theta)^(-5))/((cos 3theta +isin3theta)^(-2) (cos 3theta -isin 3theta)^(-9)) (iii) ((sinpi//8 +icos pi//8)^(8))/((sinpi//8-icospi//8)^(8))

Simplify : ((cos3theta+isin3theta)^7(cos5theta-isin5theta)^4)/((cos4theta+isin4theta)^10(cos13theta-isin13theta)^3)

(3+cos4 theta)cos2 theta=4(cos^(8)theta-sin^(8)theta)

If (costheta+isintheta)(cos2theta+isin2theta)…..(cosntheta+isinntheta)=1 , then the value of theta , is

In the form of A+iBz=(cos2 theta+i sin2 theta)^(-5)(cos3 theta-i sin3 theta)^(6)(sin theta-i cos theta)^(3) is