已知f(x)=sin²(x+π/4) 若a=f﹙lg5﹚,b=f(lg1/5)则 则a+b=0 a+b=1a-b=0a-b=1
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![已知f(x)=sin²(x+π/4) 若a=f﹙lg5﹚,b=f(lg1/5)则 则a+b=0 a+b=1a-b=0a-b=1](/uploads/image/z/12866344-16-4.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dsin%26%23178%3B%EF%BC%88x%2B%CF%80%EF%BC%8F4%29+%E8%8B%A5a%3Df%EF%B9%99lg5%EF%B9%9A%2Cb%3Df%EF%BC%88lg1%EF%BC%8F5%EF%BC%89%E5%88%99+%E5%88%99a%2Bb%3D0+a%2Bb%3D1a-b%3D0a-b%3D1)
已知f(x)=sin²(x+π/4) 若a=f﹙lg5﹚,b=f(lg1/5)则 则a+b=0 a+b=1a-b=0a-b=1
已知f(x)=sin²(x+π/4) 若a=f﹙lg5﹚,b=f(lg1/5)则 则
a+b=0
a+b=1
a-b=0
a-b=1
已知f(x)=sin²(x+π/4) 若a=f﹙lg5﹚,b=f(lg1/5)则 则a+b=0 a+b=1a-b=0a-b=1
f(x)=sin²(x+π/4)=(1-cos(2x+π/2))/2=(1+sin2x)/2
2f(x)-1=sin2x
函数是奇函数
所以
2f(x)-1+2f(-x)-1=0
f(x)+f(-x)=1
a=f(lg5),b=f(lg1/5)=f(-lg5)
所以
a+b=1
直觉告诉我,a+b=1,呵呵