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F(x)=2分之根号2cos(2x+4分之兀)+sin2x (1)求最小正周期
人气:105 ℃ 时间:2020-04-16 11:48:41
解答
f(x)=√2/2cos(2x+π/4)+sin(2x)
=cos(π/4)cos(2x+π/4)+sin(2x)
=1/2cos(2x+π/2)+1/2cos(2x)+sin(2x)
=1/2sin(2x)+1/2cos(2x)
=√2/2sin(2x+π/4)
0≤2x+π/4≤2π
-π/8≤x≤7π/8
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