3(sin 0 - cos 0)4 + 6(sin e + cos 0)2 +4(sinº @ + coso e) = 13.
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Answer:
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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