两道高一数学三角函数题

2025-01-31 10:32:26
推荐回答(2个)
回答1:

(1)见楼上,(2)
sinx+cosx=a
sinxcosx=a平方a^2-2a-1=0,a=1+根2(舍,sinx+cosx的取值)或1-根2
(sinx+cosx)(1-sinxcosx)=a(1-a)=根2-2
tanx+cotx=1/(sinxcosx)=1/a=-根2-1(通分)

回答2:

(1)sin^6A+cos^6A=(sin^2A+cos^2A)(sin^4A+cos^4A-sin^2Acos^2A)=sin^4A+cos^4A+2sin^2Acos^2A-32sin^2Acos^2A
=(sin^2A+cos^2A)^2-3sin^2Acos^A
=1-3sin^2Acos^A