问题
解答题
已经函数f(x)=2sinxcosx+sin2x-cos2x. (1)求f(x)递增区间; (2)求f(x)当x∈[0,
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答案
∵f(x)=2sinxcosx+sin2x-cos2x
=sin2x-cos2x
=
(cos(-2
)sin2x+sin(-π 4
)cos2x)π 4
=
sin(2x-2
)π 4
(1)f(x)递增区间为2x-
∈[-π 4
+2kπ,π 2
+2kπ] k∈Zπ 2
即递增区间为x∈[-
+kπ,π 8
+kπ]k∈Z)3π 8
(2)当x∈[0,
]π 2
即2x-
∈[-π 4
,π 4
]3π 4
∴f(x)min=
sin(-2
)=-1π 4
f(x)max=
sin(2
)=π 2 2
即f(x)当x∈[0,
]时的值域为[-1,π 2
]2