If pn = cosnθ + sinnθ, then pn - pn-2 = kpn-4, where :

Question

If pn = cosnθ + sinnθ, then pn - pn-2 = kpn-4, where : (A) k = 1 (B) k = -sin2 θ cos2 θ  (C) k = sin2 θ  (D) k = cos2 θ

Tardigrade Admin · Accepted Answer

pn - pn-2 = (cosn θ + sinn θ)-(cosn-2θ + sinn-2 θ) = cosn-2θ (cos2θ - 1) + sinn-2 θ (sin2 θ - 1 ) = -sin2 θ cosn-2 θ - cos2 θ sinn-2 θ = -sin2 θ cos2 θ (cosn-4 θ + sinn-4 θ) = -sin2 θ cos2 θ pn-4 = kpn-4 ⇒ k = -sin2 θ cos2 θ

Q. If $p_{n} = co s^{n} θ + s i n^{n} θ$ , then $p n - p_{n - 2} = k p_{n - 4}$ , where :

$k = 1$

$k = - s i n^{2} θ co s^{2} θ$

$k = s i n^{2} θ$

$k = co s^{2} θ$

Q. If pn​=cosnθ+sinnθ, then pn−pn−2​=kpn−4​, where :

k=1

k=−sin2θcos2θ

k=sin2θ

k=cos2θ

pn​−pn−2​=(cosnθ+sinnθ)−(cosn−2θ+sinn−2θ) =cosn−2θ(cos2θ−1)+sinn−2θ(sin2θ−1) =−sin2θcosn−2θ−cos2θsinn−2θ =−sin2θcos2θ(cosn−4θ+sinn−4θ) =−sin2θcos2θpn−4​=kpn−4​ ⇒k=−sin2θcos2θ

Q. If $p_{n} = co s^{n} θ + s i n^{n} θ$ , then $p n - p_{n - 2} = k p_{n - 4}$ , where :

$k = 1$

$k = - s i n^{2} θ co s^{2} θ$

$k = s i n^{2} θ$

$k = co s^{2} θ$