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प्रश्न
If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]
उत्तर
LHS = \[6 T_{10} - 15 T_8 + 10 T_6 - 1\]
\[=6\left( \sin^{10} x + \cos^{10} x \right) - 15\left( \sin^8 x + \cos^8 x \right) + 10\left( \sin^6 x + \cos^6 x \right) - 1\]
`=6(sin^2x+cos^2x)(sin^8x+cos^8x-sin^2xcos^2x)-15(sin^8x+cos^8x)+10(sin^6x+cos^6x)-1`
`=6(sin^8x+cos^8x-sin^2xcos^2x)-15(sin^8x+cos^8x)+10(sin^6x+cos^6x)-1`
`=6sin^8x+6cos^8x-6sin^2xcos^2x-15sin^8x-15cos^8x+10(sin^6x+cos^6x)-1`
`=-6sin^2xcos^2x-9sin^8x-9cos^8x+10(sin^6x+cos^6x)-1`
`=-6sin^2xcos^2x-9(sin^8x+cos^8x)+10(sin^6x+cos^6x)-1`
`=-6sin^2xcos^2x-9(sin^2x+cos^2x)(sin^6x+cos^6x-sin^2xcos^2x)+10(sin^6x+cos^6x)-1`
`=-6sin^2xcos^2x-9(sin^6x+cos^6x-sin^2xcos^2x)+10(sin^6x+cos^6x)-1`
`=-6sin^2xcos^2x-9sin^6x-9cos^6x+9sin^2xcos^2x+10sin^6x+10cos^6x-1`
`=3sin^2xcos^2x+sin^6x+cos^6x-1`
`=3sin^2xcos^2x+(sin^2x+cos^2x)(sin^4x+cos^4x-sin^2xcos^2x)-1`
`=3sin^2xcos^2x+sin^4x+cos^4x-sin^2xcos^2x-1`
`=(sin^2x)^2+2sin^2xcos^2x+(cos^2x)^2-1`
`=(sin^2x+cos^2x)^2-1`
=12-1
=0
=RHS
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