If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3} \]

Solution:

\[ T_3-T_5 = (\sin^3 x+\cos^3 x) -(\sin^5 x+\cos^5 x) \]

\[ = \sin^3 x(1-\sin^2 x) + \cos^3 x(1-\cos^2 x) \]

\[ = \sin^3 x\cos^2 x + \cos^3 x\sin^2 x \]

\[ = \sin^2 x\cos^2 x(\sin x+\cos x) \]

\[ = \sin^2 x\cos^2 x\;T_1 \]

Therefore,

\[ \frac{T_3-T_5}{T_1} = \sin^2 x\cos^2 x \]

Similarly,

\[ T_5-T_7 = \sin^5 x(1-\sin^2 x) + \cos^5 x(1-\cos^2 x) \]

\[ = \sin^5 x\cos^2 x + \cos^5 x\sin^2 x \]

\[ = \sin^2 x\cos^2 x(\sin^3 x+\cos^3 x) \]

\[ = \sin^2 x\cos^2 x\;T_3 \]

Hence,

\[ \frac{T_5-T_7}{T_3} = \sin^2 x\cos^2 x \]

Therefore,

\[ \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3} \]

Hence proved.