If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ 2T_6-3T_4+1=0 \]

Solution:

\[ T_6=\sin^6 x+\cos^6 x \]

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

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

Also,

\[ T_4=\sin^4 x+\cos^4 x \]

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

\[ = 1-2\sin^2 x\cos^2 x \]

Now,

\[ 2T_6-3T_4+1 \]

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

\[ = 2-6\sin^2 x\cos^2 x -3+6\sin^2 x\cos^2 x +1 \]

\[ =0 \]

Hence proved.