设f(x)是以2为周期的周期函数,又是奇函数,且f(-2/5)=3,sina=√5/5,则f(4cos2a)的值是
问题描述：
设f(x)是以2为周期的周期函数,又是奇函数,且f(-2/5)=3,sina=√5/5,则f(4cos2a)的值是
怎么做的,请详细解析
答
cos2a = 1 - 2*(sina)^2 = 1 - 2 * (√5/5)^2
= 1 - 2 * 1/5
= 3/5
4*cos2a = 4*3/5 = 12/5
f(x)是以2为周期的周期函数,所以
f(12/5) = f(2 + 2/5) = f(2/5)
f(x) 又是奇函数 ,所以
f(-x) = - f(x)
f(2/5) = - f(-2/5) = -3
f(4cos2a)的值是 -3