Question

数学三角函数题目

已知f(X)=(sin2X+cos2X)/(tanX+cotX).求函数的最小周期。以及值域

Answer 1

点击下图

Answer 2

f(x)=(sin2x+cos2x)/(tanx+cotx)
=(sin2x+cos2x)/(sinx/cosx+cosx/sinx)
=(sin2x+cos2x)/[(si'n^x+cos^x)/sinxcosx]
=(sin2x+cos2x)/(1/sinxcosx)
=(sin2x+cos2x)sinxcosx
=(1/2)(sin2x+cos2x)sin2x
=(1/4)[2sin^(2x)+2cos2xsin2x]
=(1/4)[1-cos4x+sin4x]
=1/4-根号2*sin(4x-∏/4]/4 ,x≠k∏/2, 4x≠2k∏
-1<=sin(4x-∏/4]<=1
根号2/4<=-根号2*sin(4x-∏/4]/4<=根号2/4,
所以(1-根号2)/4<=1/4-根号2*sin(4x-∏/4]/4<=(1+根号2)/4
所以f(x)的值域[(1-根号2)/4,(1+根号2)/4]
周期2∏/4=∏/2