If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ 6T_{10}-15T_8+10T_6-1=0 \]

Solution:

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

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

\[ T_8=\sin^8 x+\cos^8 x \]

\[ = (\sin^4 x+\cos^4 x)^2-2\sin^4 x\cos^4 x \]

\[ = (1-2\sin^2 x\cos^2 x)^2 -2\sin^4 x\cos^4 x \]

\[ = 1-4\sin^2 x\cos^2 x+2\sin^4 x\cos^4 x \]

\[ T_{10}=\sin^{10} x+\cos^{10} x \]

\[ = (\sin^2 x+\cos^2 x)^5 -5\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)^3 \]

\[ \qquad +5\sin^4 x\cos^4 x(\sin^2 x+\cos^2 x) \]

\[ = 1-5\sin^2 x\cos^2 x +5\sin^4 x\cos^4 x \]

Now,

\[ 6T_{10}-15T_8+10T_6-1 \]

\[ = 6(1-5s+5s^2) -15(1-4s+2s^2) +10(1-3s)-1 \]

where \[ s=\sin^2 x\cos^2 x \]

\[ = 6-30s+30s^2 -15+60s-30s^2 +10-30s-1 \]

\[ =0 \]

Hence proved.