# If u_n=sin^n(theta)+cos^n(theta) and (u_3-u_5)/(u_1)=K((u_5-u_7)/(u_3)) then find K?

Jul 21, 2018

$K = 1$.

#### Explanation:

Given that, ${u}_{n} = {\sin}^{n} \theta + {\cos}^{n} \theta$.

$\therefore {u}_{3} - {u}_{5} = \left({\sin}^{3} \theta + {\cos}^{3} \theta\right) - \left({\sin}^{5} \theta + {\cos}^{5} \theta\right)$,

$= \left({\sin}^{3} \theta - {\sin}^{5} \theta\right) + \left({\cos}^{3} \theta - {\cos}^{5} \theta\right)$,

$= {\sin}^{3} \theta \left(1 - {\sin}^{2} \theta\right) + {\cos}^{3} \theta \left(1 - {\cos}^{2} \theta\right)$,

$= {\sin}^{3} \theta {\cos}^{2} \theta + {\cos}^{3} \theta {\sin}^{2} \theta$,

$= {\sin}^{2} \theta {\cos}^{2} \theta \left(\sin \theta + \cos \theta\right)$,

$= {u}_{1} \left({\sin}^{2} \theta {\cos}^{2} \theta\right)$.

$\Rightarrow \frac{{u}_{3} - {u}_{5}}{u} _ 1 = {\sin}^{2} \theta {\cos}^{2} \theta \ldots \ldots \ldots . \left({\ast}^{1}\right)$.

Again, ${u}_{5} - {u}_{7} = {\sin}^{5} \theta + {\cos}^{5} \theta - {\sin}^{7} \theta - {\cos}^{7} \theta$,

$= {\sin}^{5} \theta \left(1 - {\sin}^{2} \theta\right) + {\cos}^{5} \theta \left(1 - {\cos}^{2} \theta\right)$,

$= {\sin}^{5} \theta {\cos}^{2} \theta + {\cos}^{5} \theta {\sin}^{2} \theta$,

$= {\sin}^{2} \theta {\cos}^{2} \theta \left({\sin}^{3} \theta + {\cos}^{3} \theta\right)$,

$= {u}_{3} \left({\sin}^{2} \theta {\cos}^{2} \theta\right)$.

$\Rightarrow \frac{{u}_{5} - {u}_{7}}{u} _ 3 = {\sin}^{2} \theta {\cos}^{2} \theta \ldots \ldots \ldots \left({\ast}^{2}\right)$.

$\text{Hence, } \frac{{u}_{3} - {u}_{5}}{u} _ 1 = K \left(\frac{{u}_{5} - {u}_{7}}{u} _ 3\right) , \left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right) ,$

$\Rightarrow K = 1$.

$\textcolor{red}{\text{Enjoy Maths.!}}$