If T_n= ^nx+ ^nx, Prove that 2 T_6 - 3 T_4 + 1 = 0

Question

If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that
$2 \text{T}_6 - 3\text{ T}_4 + 1 = 0$

✓

Answer

$\text{L.H.S}=2\text{T}_6-3\text{T}_4+1$
$=2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1$
$=2((\sin^2\text{x})^3+(\cos^2\text{x})^3-2(\sin^2\text{x})^2+(\cos^2\text{x})^2)+1$
$=2((\sin^2\text{x}+\cos^2\text{x})(\sin^2\text{x})^2+(\cos^2\text{x})^2-(\sin^2\text{x}\cos^2\text{x}))\\\ \ \ -3((\sin^2\text{x})^2+(\cos^2\text{x})^2+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})=1$ $ [$Using $a^3 + b^3 =(a + b)(a^2 + b^2 - ab)$ and adding and subtracting $2\sin^2\text{x}\cos^2\text{x}\big]$
$=2((\sin^2\text{x}+\cos^2\text{x})^2-3\sin^2\text{x}\cos^2\text{x})-3(1-2\sin^2\text{x}\cos^2\text{x})+1$
$=2(1-3\sin^2\text{x}\cos^2\text{x})-3\sin^2\text{x}\cos^2\text{x}+1$
$=2-6\sin^2\text{x}\cos^2\text{x}-2+6\sin^2\text{x}\cos^2\text{x}$
$=0$
$=\text{R.H.S}$
$\text{Proved}.$

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