( sin θ +i cos θ )n is equal to

Question

( sin θ +i cos θ )n is equal to (A)  in(i sin θ + cos θ )  (B)  ( sin nθ +i cos nθ )  (C)  cos n( (π /2)-θ )+i sin n( (π /2)-θ )  (D)  cos θ +i sin nθ

Tardigrade Admin · Accepted Answer

( sin θ +i cos θ )hn =[ cos ( (π /2)-θ )+i sin ( (π /2)-θ ) ]n = cos n( (π /2)-θ )+i sin n( (π /2)-θ )

Q. $(sin θ + i cos θ)^{n}$ is equal to

$i^{n} (i sin θ + cos θ)$

$(sin n θ + i cos n θ)$

$cos n (\frac{π}{2} - θ) + i sin n (\frac{π}{2} - θ)$

$cos θ + i sin n θ$

$(sin θ + i cos θ) hn$
$= [cos (\frac{π}{2} - θ) + i sin (\frac{π}{2} - θ)]^{n}$
$= cos n (\frac{π}{2} - θ) + i sin n (\frac{π}{2} - θ)$

Q. (sinθ+icosθ)n is equal to

in(isinθ+cosθ)

(sinnθ+icosnθ)

cosn(2π​−θ)+isinn(2π​−θ)

cosθ+isinnθ