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Question
Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]
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Solution
\[LHS = 3 \left( \text{ sin } x - \text{ cos } x \right)^4 + 6 \left( \text{ sin } x + \text{ cos } x \right)^2 + 4\left( \sin^6 x + \cos^6 x \right)\]
`3{(sinx-cosx)^2}^2+6(sin^2x+cos^2x+2sinx.cosx)+4{(sin^2x)^3+(cos^2)^3}`
= `3(sin^2x+cos^2x-2sinxcosx)^2+6(1+2sinxcosx)+4{(sin^2x+cos^2x)((sin^2x)^2+(cos^2x)^2-sin^2x+cos^2x)}` ...`["a"^3+"b"^3=("a"+"b")("a"^2+"b"^2-"ab")]`
= `3(1-2sinxcosx)^2+6(1+2sinxcosx)+4(sin^4x+cos^4-sin^2xcos^2x)`
= `3(1+4sin^2xcos^2x-4sinxcosx)+6+12sinxcosx+4{(sin^2x+cos^2x)^2-2cos^2xsin^2x-sin^2xcos^2x}`
= `3+12sin^2cos^2x-12sinxcosx+6+12sinxcosx+4{(sin^2x+cos^2)^2-sin^2xcos^2x}`
= `9+12sin^2cos^2x+4-12sin^2cos^2x`
= 13 = R.H.S
Hence proved.
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