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

1.答案为：【根号2/2,根号2】
y=-2sin(π/4)*sin(-π/6-x) ,(和差化积的公式)
则y=根号2*sin(π/6+x),由x∈[0,π/2],则x+π/6∈[π/6,2π/3],
由sinx的图像可得：y∈【根号2/2,根号2】
2.f(x)=根号(a^2+b^2)sin(x+v),其中tanv=b/a,所以有2=根号(a^2+b^2)
即a^2+b^2=4,且f（π/6)=a/2+根号3b/2=根号3
则由这两个式子可求得：a=根号3,b=1或a=0,b=2（舍去）
所以f（π/3)=2
3.由1+sin10°=1+2sin5°cos5°=(sin5°+cos5°)^2 同理1-sin10°=(sin5°-cos5°)^2
又因为在x∈[0,π/4]使有cosx>sinx
则原式=sin5°+cos5°+cos5°-sin5°=2cos5°

y=cos(π/12- x)-cos(5π/12+x)
=cos(π/2-5π/12- x)-cos(5π/12+x)
=sin(5π/12+ x)-cos(5π/12+x)
=√2*[√2/2sin(5π/12+ x)-√2/2cos(5π/12+x)]
=√2*[sin(5π/12+ x)cosπ/4-√2/2(5π/12+x)sinπ/4]
=√2*s...

1. 已知x∈[0，π/2],求函数y=cos（π/12 - x）-cos（5π/12 + x）的值域？
利用：cosA -cosB = -2sin(A+B/2)sin(A-B/2)
y=cos（π/12 - x）-cos（5π/12 + x）＝ - 2 cos(π/4)cos(x +π/6 ) ＝ -√2cos(x +π/6 )
y的值域为： - √2 ...

第1题和差化积y=-1/2sin(π/12 - x+5π/12 + x)/2sin(π/12 - x-5π/12 - x)/2
√(1+2sin5cos5)+√(1-2sin5cos5)
√(sin5+cos5)2+√(sin5-cos5)2
sin5+cos5+sin5-cos5
2sin5