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Question

Prove that 3(sinxcosx)4+4(sin6x+cos6x)+6(sinx+cosx)2=13

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Solution

Consider 4(sin6x+cos6x)
=4[(sin2x)3+(cos2x)3]
=4[(sin2x+cos2x)(sin4xsin2xcos2x+cos4x)]
=4[(sin2x+cos2x)22sin2xcos2x2sin2xcos2x]
=4[13sin2xcos2x]
=412sin2xcos2x ........(1)
6[sinx+cosx]2
=6[sin2x+cos2x+2sinxcosx]
=6[1+2sinxcosx]
=6+12sinxcosx ......(2)
3(sinxcosx)4
=3[(sinxcosx)2]2
=3[sin2x+cos2x2sinxcosx]2
=3[12sinxcosx]2
=3[14sinxcosx+4sin2xcos2x]
=312sinxcosx+12sin2xcos2x .....(3)
Adding (1),(2) and (3) we get
3(sinxcosx)4+4(sin6x+cos6x)+6[sinx+cosx]2
=312sinxcosx+12sin2xcos2x+412sin2xcos2x+6+12sinxcosx
=13
Hence proved

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