Question

Prove that:
3 (sin x - cos x)^4 + 4 (sin^6×x + cos^6×x) + 6 (sin x + cos x)^2 = 13.​

Answer 1

Step-by-step explanation:

Consider 4(sin

6

x+cos

6

x)

=4[(sin

2

x)

3

+(cos

2

x)

3

]

=4[(sin

2

x+cos

2

x)(sin

4

x−sin

2

xcos

2

x+cos

4

x)]

=4[(sin

2

x+cos

2

x)

2

−2sin

2

xcos

2

x−2sin

2

xcos

2

x]

=4[1−3sin

2

xcos

2

x]

=4−12sin

2

xcos

2

x ........(1)

6[sinx+cosx]

2

=6[sin

2

x+cos

2

x+2sinxcosx]

=6[1+2sinxcosx]

=6+12sinxcosx ......(2)

3(sinx−cosx)

4

=3[(sinx−cosx)

2

]

2

=3[sin

2

x+cos

2

x−2sinxcosx]

2

=3[1−2sinxcosx]

2

=3[1−4sinxcosx+4sin

2

xcos

2

x]

=3−12sinxcosx+12sin

2

xcos

2

x .....(3)

Adding (1),(2) and (3) we get

3(sinx−cosx)

4

+4(sin

6

x+cos

6

x)+6[sinx+cosx]

2

=3−12sinxcosx+12sin

2

xcos

2

x+4−12sin

2

xcos

2

x+6+12sinxcosx

=13

Hence proved

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