Question

高一数学~

f（x）=sin2x+cos2x/cos2x-sin2x 求最小正周期~

Answer 1

f（x）=（sin2x+cos2x）/（cos2x-sin2x ）
=√2sin(2x+π/4)/[√ 2cos(2x+π/4)]
=sin(2x+π/4)/[cos(2x+π/4)]
=tan(2x+π/4)
f（x）最小正周期
T=π/2

追问

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追答

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追问

看不懂~

追答

sin2x+cos2x=√2(√2/2*sin2x+√2/2*cos2x)
=√2(cosπ/4*sin2x+sinπ/4*cos2x)
=√2sin(2x+π/4)
cos2x-sin2x =√ 2cos(2x+π/4)方法同上

Answer 2