Question

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

• A. True
• B. False

Answer 1

The correct option is B False

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

$T_n-T_{n-2}={\sin }^n\theta +{\cos }^n\theta -{\sin }^{n-2}\theta -{\cos }^{n-2}\theta$

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

$=-{\sin }^{n-2}\theta {\cos }^2\theta -{\cos }^{n-2}\theta {\sin }^2\theta$

$=-{\sin }^2\theta {\cos }^2\theta ({\sin }^{n-4}\theta +{\cos }^{n-4}\theta )$

$T_{n-2}-T_n={\sin }^2\theta {\cos }^2\theta T_{n-4}$

So $\frac{T_{n-2}-T_n}{T_{n-4}}={\sin }^2\theta {\cos }^2\theta$

$\frac{T_3-T_5}{T_1}={\sin }^2\theta {\cos }^2\theta$

$\frac{T_5-T_7}{T_3}={\sin }^2\theta {\cos }^2\theta$

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

So the given relation is $\text{False}$