Question

If $T_n={\sin }^{n}x+{\cos }^{n}x$, then find $\frac{T_3-T_5}{T_1}-\frac{T_5-T_7}{T_3}$.

Answer 1

$T_n={\sin }^{n}x+{\cos }^{n}x$

$T_3={\sin }^{3}x+{\cos }^{3}x$

$T_4={\sin }^{4}x+{\cos }^{4}x$

$T_5={\sin }^{5}x+{\cos }^{5}x$

$T_6={sin}^{6}x+{cos}^{6}x$

similarly we can find $T_1,T_7$ also,

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

$\left(I\right)$ $\left(II\right)$

$\left(I\right)$ $=\frac{{\sin }^{3}x+{\cos }^{3}x-{\sin }^{5}x-{\cos }^{5}x}{\sin x+\cos x}$

$\Rightarrow \frac{{\sin }^{3}x(1-{\sin }^{2}x)+{\cos }^{3}x(1-{\cos }^{2}x)}{\sin x+\cos x}$

$\because {\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow \frac{{\sin }^{3}x.\left({\cos }^{2}x\right)+{\cos }^{3}x\left({\sin }^{2}x\right)}{\sin x+\cos x}$

$\Rightarrow \frac{{\sin }^{2}x{\cos }^{2}x(\sin x+\cos x)}{\sin x+\cos x}$

$I={\sin }^{2}x{\cos }^{2}x⟶\left(1\right)$

$II=\frac{T_5-T_7}{T_3}$

$\Rightarrow \frac{{\sin }^{5}x+{\cos }^{5}x-{\sin }^{7}x-{\cos }^{7}x}{{\sin }^{3}x+{\cos }^{3}x}$

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

$\because {\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow \frac{{\sin }^{5}x{\cos }^{2}x+{\cos }^{5}x{\sin }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}$

$\Rightarrow \frac{{\sin }^{2}x{\cos }^{2}x({\sin }^{3}x+{\cos }^{3}x)}{{\sin }^{3}x+{\cos }^{3}x}$

$\Rightarrow {\sin }^{2}x{\cos }^{2}x=II$

So $I=II$

$\Rightarrow I-II=0$