已知x∈[0,π/2],求函数y=cos（π/12 - x）-cos（5π/12 + x）的值域?已知函数f（x）=a*sinx+b*cosx,（ab≠0）的最大值是2,且f（π/6)=根号3.求f（π/3)化简:根号【1+sin10°】+根号【1-sin10°】

已知x∈[0,π/2],求函数y=cos（π/12 - x）-cos（5π/12 + x）的值域?已知函数f（x）=a*sinx+b*cosx,（ab≠0）的最大值是2,且f（π/6)=根号3.求f（π/3)化简:根号【1+sin10°】+根号【1-sin10°】
已知x∈[0,π/2],求函数y=cos（π/12 - x）-cos（5π/12 + x）的值域?
已知函数f（x）=a*sinx+b*cosx,（ab≠0）的最大值是2,且f（π/6)=根号3.求f（π/3)
化简:根号【1+sin10°】+根号【1-sin10°】

已知x∈[0,π/2],求函数y=cos（π/12 - x）-cos（5π/12 + x）的值域?已知函数f（x）=a*sinx+b*cosx,（ab≠0）的最大值是2,且f（π/6)=根号3.求f（π/3)化简:根号【1+sin10°】+根号【1-sin10°】
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°

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 ...

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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 ×[ -1/2, √3/2 ]
我也用到了f(x)在x∈[0，π/2]的范围内单调减小的
2. 已知函数f（x）=a*sinx+b*cosx，（ab≠0）的最大值是2，且f（π/6)=根号3.求f（π/3)
f（x）=a*sinx+b*cosx = √(a²+b²)× [a/√(a²+b²) sinx + b/√(a²+b²)cosx ]
= √(a²+b²)×[cosM sinx+ sinM cosx] --------------------这里设sinM = b/√(a²+b²)
= √(a²+b²)× sin(x+M)
⑴ 最大值f（x）=√(a²+b²)× 1 = 最大值是2 ====>√(a²+b²) = 2
⑵ f（π/6)=√3 ====> √(a²+b²)× sin(π/6 + M) =√3 ====> M = π/6
f（π/3) = √(a²+b²)× sin(x+M) = 2sin(π/3+π/6) = 2
3. 化简:根号【1+sin10°】+根号【1-sin10°】
原式 = 根号【1+sin10°】+根号【1-sin10°】 ----------为了方便：设Sin(5°) = M
=√(1+sin10°) + √(1-sin10°) ------------- sin(2A) = 2sinAcosA； 1 =sin² +cos ²
= √(sin²5°+ 2sin5°cos5°+ cos²5°) + √(sin²5°- 2sin5°cos5°+ cos²5°)
=√ (sin5°+ cos5°) ² + √ (sin5°- cos5°) ² sin5°< cos5°
= (sin5°+ cos5°) + ( cos5°-sin5°)
= 2cos5°
呵，楼主很有魅力，吸引我门都来了，我最慢了，过程又最麻烦，感谢前面两位吧
三道题求教,过程不用很详细，能看懂就ok~！谢谢了！！

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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...

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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*sin(5π/12+ x-π/4)
=√2*sin(5π/12+ x-π/4)
=√2*sin(π/6+ x)
0<=π/6+ x<=π/2
-π/6<= x<=π/3
-1/2<=sin(π/6+ x)<=√3/2
-√2/2<=√2sin(π/6+ x)<=√6/2
-√2/2<=y<=√6/2
f（x）=a*sinx+b*cosx
=√(a^2+b^2)sin(A+x)(其中tanA=b/a)
√(a^2+b^2)=2
a^2+b^2=2
f(π/6)=√(a^2+b^2)sin(A+π/6)
f(π/6)=2sin(A+π/6)
2sin(A+π/6)=√3
sin(A+π/6)=√3/2
A+π/6=π/3或A+π/6=2π/3
A=π/6或A=π/2（舍去）
所以A=π/6
f(π/3)
=2sin(π/6+π/3)
=2sin(π/2)
=2
√(1+sin10°)+√(1-sin10°)
=√[(sin5°)^2+2sin5°cos5°+(cos5°)^2]+√[(sin5°)^2-2sin5°cos5°+(cos5°)^2]
=√(sin5°+cos5°)^2+√(sin5°-cos5°)^2 (sin5°=sin5°+cos5°+cos5°-sin5°
=2cos5°

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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