已知函数 f(x)=sin(x+ 7π 4 )+cos(x- 3π 4 ) ,x∈R,求f(x)的最小正周期和在[0,

2025-06-27 18:13:57
推荐回答(1个)
回答1:

f(x)=sinxcos
4
+cosxsin
4
+cosxcos
4
+sinxsin
4

=
2
2
sinx-
2
2
cosx-
2
2
cosx+
2
2
sinx
=
2
(sinx-cosx)
=2sin(x-
π
4
),
∵ω=1,∴T=2π;
∵x∈[0,
π
2
],∴x-
π
4
∈[-
π
4
π
4
],
∴-
2
2
≤sin(x-
π
4
)≤
2
2
,即-
2
≤2sin(x-
π
4
)≤
2

则函数在[0,
π
2
]上的最大值为
2
,最小值为-
2