`((costheta+isintheta)/(sintheta+icostheta))^4=cos8theta+isin8theta`

Prove that (frac(1+costheta+isintheta)(1+costheta-isintheta))^n=cos ntheta+isin ntheta .

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

Simplify: cos theta[{:(costheta,sintheta),(-sintheta,costheta):}]+sintheta[{:(sin theta ,-costheta),(costheta, sintheta):}]

(sin3theta-cos3theta)/(sintheta+costheta)+1 =

(sintheta+sin2theta)/(1+costheta+cos2theta) =

(sintheta+sin2theta)/(1+costheta+cos2theta)=?

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

The general value of theta which satisfies the equation (costheta+isintheta)(cos3theta+isin3theta) (cos5theta+isin5theta)…………((cos2n-1)theta+isin(2n-1)theta) = 1 is

Show that costheta [[costheta, sintheta],[-sintheta, cos theta]] + sin theta [[sintheta ,-cos theta],[costheta, sin theta]] =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))