Question

为什么函数y=cos^2(x-∏/12）+sin^2(x+∏/12)-1是 奇函数

Answer 1

y=cos²(x-π/12) + sin²(x＋π/12)－1
=(1+cos(2x-π/6))/2+(1-cos(2x+π/6))/2－1
= cos(2x-π/6)/2-cos(2x+π/6)/2
=1/2[cos(2x-π/6)- cos(2x+π/6)]
=1/2[cos2x cosπ/6+sin2x sinπ/6-( cos2x cosπ/6-sin2x sinπ/6)]
=1/2[2sin2x sinπ/6]
= sin2x sinπ/6
=1/2 sin2x
∴函数的最小正周期是π
即周期为π的奇函数

Answer 2

y=cos^2(x-∏/12）+sin^2(x+∏/12)-1
=cos^2(x-∏/12）-cos^2(x+∏/12)
=[cos(x-∏/12)+cos(x+∏/12)][cos(x-∏/12)-cos(x+∏/12)]
=[2cosxcos(∏/12)][2sinxsin(∏/12)]
=[2cosxsinx][2cos(∏/12)sin(∏/12)]
=[2cosxsinx][sin(∏/6)]
=cosxsinx,
令f(x)=cosxsinx,则f(-x)=cos(-x)sin(-x)=-cosxsinx=-f(x),
所以y=cos^2(x-∏/12）+sin^2(x+∏/12)-1=cosxsinx是奇函数

Answer 3

f(x)=cos^2(x-π/12）+sin^2(x+π/12)-1

f(-x)=cos^2(-x-π/12）+sin^2(-x+π/12)-1
=cos^2(x+π/12）+sin^2(x-π/12)-1
=1-sin^2(x+π/12)+1-cos^2(x-π/12）-1
=1-sin^2(x+π/12)+1-cos^2(x-π/12）
=-(cos^2(x-π/12）+sin^2(x+π/12)-1)
=-f(x)
所以是奇函数