Question

If $T_n={\sin }^n\theta +{\cos }^n\theta$, prove that $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}$

Answer 1

Given, $T_m={\sin }^n\theta +{\cos }^n\theta$

$\frac{T_3-T_5}{T_1}=\frac{{\sin }^3\theta -{\sin }^5\theta +{\cos }^3\theta -{\cos }^5\theta }{\sin \theta +\cos \theta }$

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

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

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

$\frac{T_5-T_7}{T_3}=\frac{{\sin }^5\theta -{\sin }^7\theta +{\cos }^5\theta -{\cos }^7\theta }{{\sin }^3\theta +{\cos }^3\theta }$

$=\frac{{\sin }^5\theta (1-{\sin }^2\theta )+{\cos }^5\theta (1-{\cos }^2\theta )}{{\sin }^3\theta +{\cos }^3\theta }$

$=\frac{{\sin }^5\theta {\cos }^2\theta +{\cos }^5\theta {\sin }^2\theta }{{\sin }^3\theta +{\cos }^3\theta }$

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

$\therefore$ From $\left(1\right)$ and $\left(2\right)$

$\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}----Proved$