1.已知x∈[0,∏/2],求函数y=cos(∏/12 -x)-cos(5∏/12 +x)的值域

1.y=cos(π/12-x)-cos(5π/12+x)
=sin(5π/12+x)-cos(5π/12+x)=√2sin(x+π/6)
y∈[√2/2,√2]
2.由sina+cosa=-1/5,且0≤a≤π,能算出sina=3/5,cosa=-4/5
所以tana=-3/4
3.y=(sinx+1)/(cosx+2)
={[1+tan(x/2)]/[1+tan^2(x/2)]+1}/{[1-tan^2(x/2)]/[1+tan^2(x/2)]+2}
=[1+tan(x/2)+1+tan^2(x/2)]/[1-tan^2(x/2)+2+2tan^2(x/2)]
=[tan(x/2)+tan^2(x/2)+2]/[tan^2(x/2)+3]
设x=tan^2(x/2)
y=(y^2+y+2)/(y^2+3)
=1+(y-1)/(y^2+3)
4.(1-cos2a+sin2a)/(1+cos2a+sin2a)
=(2sin^2a+2sinacosa)/(2cos^2a+2sinacosa)
=tana
5.