Question

函数y=2sinx的平方+2sinxcosx的最小正周期

Answer 1

2倍角公式
cos2x=cos²x-sin²x=1-2sin²x 2sin²x=1-cos2x
sin2x=2sinxcosx
y=2sin²x+2sinxcosx
=1-cos2x+sin2x
=sin2x-cos2x+1
=√2(√2/2sin2x-√2/2cos2x)+1
=√2(cosπ/4sin2x-sinπ/4cos2x)+1
=√2sin(2x-π/4)+1
T=2π/2=π
最小正周期π
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Answer 2

y=2sin²x+2sinxcosx
=1-cos2x+sin2x
=√2sin(2x-π/4)+1
所以最小正周期是T=2π/2=π

Answer 3

2sinxcosx=sin(2x);
2sin方x=2-2cos方(x)=1-cos(2x);
所以，原式=1+sin(2x)-cos(2x)=1+根号2*sin(2x-π/4)
所以f(2x)的最小正周期=2π,f(x)的最小正周期=π

Answer 4

=1-cos2x+sin2x
=1+√2(√2/2sin2x-√2/2cos2x)
=1+√2sin(2x+4/TT)
所以最小正周期T=2TT/2=TT

Answer 5

y=2sinx的平方+2sinxcosx
=1-cos2x+sin2x
=根号2 sin(2x-π/4)+1
T=π