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Question
If \[T_n = \sin^n x + \cos^n x\], prove that \[2 T_6 - 3 T_4 + 1 = 0\]
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Solution
LHS: \[2 T_6 - 3 T_4 + 1\]
\[2\left( \sin^6 x + \cos^6 x \right) - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[2\left( \sin^2 x + \cos^2 x \right)\left( \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x \right) - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[2 . 1 . \left( \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x \right) - 3\left( \sin^4 x + \cos^4 x \right) + 1\]
\[2 \sin^4 x + 2 \cos^4 x - 2 \sin^2 x \cos^2 x - 3 \sin^4 x - 3 \cos^4 x + 1\]
\[ - \left( \sin^4 x + \cos^4 x \right) - \sin^2 x \cos^2 x + 1\]
\[ - ( \sin^2 x + \cos^2 x )^2 + 1\]
\[ - 1 + 1\]
\[0\]
Hence proved.
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