Question

高一数学？

函数y=cos²（x-π/12)+sin²(x+π/12)-1的最小正周期和奇偶性？？

Answer 1

手机不能写“PI”我用P表示吧！
cos^2(x-P/12)+sin^2(x+P/12)-1
=[cos(2x-P/6)+1+1+cos(2x+P/6)]/2-1
=cos(2x-P/6)+sin(2x-P/3)
(展开)=3^(1/2)/2cos2x+1/2sin2x+1/2sin2x-3^(1/2)/2cos2x
=sin2x

所以，最小周期：T=P(就是手机打不出来的“PI”)
函数奇偶性是：奇函数

请满意！＞_＜
手酸拉…

Answer 2

因为，cos²（x+π/12)+sin²(x+π/12)=1
则，y=cos²（x-π/12)+sin²(x+π/12)-[cos²（x+π/12)+sin²(x+π/12)]
=cos²（x-π/12)-cos²（x+π/12)
=[cos（x-π/12)+cos（x+π/12)][cos（x-π/12)-cos（x+π/12)]
=2cosxcos(π/12）*2sinxsin（π/12）
=2sinxcosx*[2sin（π/12)cos(π/12)]
=sin2x*sin（π/6)
=1/2sin2x
所以，周期为π，为奇函数。

Answer 3

奇函数，y=0.5sin(2x),最小正周期∏

Answer 4

y=[cos(x-π/12)]^2+[sin(x+π/12)]^2-1
=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1
=[cos(2x-π/6)-cos(2x+π/6)]/2
=(1/2)*(-2)sin2xsin(-π/6)
=(1/2)sin2x

π 奇函数

Answer 5