问题 填空题
函数y=cos2x-3sinxsin(x+
2
)
的最小正周期T=______.
答案

y=cos2x-3sinxsin(x+

2

=

1
2
(1+cos2x)-3sinx(sinxcos
2
+cosxsin
2

=

1
2
+
1
2
cos2x+3sinxcosx

=

1
2
+
1
2
cos2x+
3
2
sin2x

=

1
2
+
10
2
10
10
cos2x+
3
10
10
sin2x)

=

1
2
+
10
2
sin(2x+θ)(其中sinθ=
10
10
,cosθ=
3
10
10
),

∵ω=2,

∴T=

2
=π.

故答案为:π

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