求救~几个高一数学题~

1.化简:sinx/(1-cosx )* √[(tanx-sinx)/(tanx+sinx)]
解：tanx=sinx/cosx
故：(tanx-sinx)/(tanx+sinx)= (sinx/cosx -sinx)/( sinx/cosx +sinx)=（1-cosx）/(1+cosx)
=（1-cosx）²/[(1+cosx) （1-cosx）]= （1-cosx）²/ sin²x
故：sinx/(1-cosx )* √[(tanx-sinx)/(tanx+sinx)]
= sinx/(1-cosx )* √[（1-cosx）²/ sin²x]
= sinx/(1-cosx )* （1-cosx）/∣sinx∣
= sinx/∣sinx∣
=±1

2.若实数x满足lgx=2+sinβ 求|x+1|-|x-10|的值域
解：因为-1≤sinβ≤1，故：1≤lgx=2+sinβ≤3
故：10≤x≤1000
当10≤x≤1000时，|x+1|-|x-10|=x+1-（x-10）=11
故：|x+1|-|x-10|的值(域)为11

3、设函数f(x)=-x²+2x+3(0≤x≤3），记f（x)的最大值m,最小值n，当角α的终边经过点P（m,n-1)时，求sinα+cosα的值
解：f(x)=-x²+2x+3=-（x-1）²+4
当0≤x≤3时，f（x)的最大值m=4（此时x=1），最小值n=0（此时x=3）
故：过点P（m,n-1)为P（4，-1）
故：|OP|=√17
故：sinα+cosα=-1/√17+4/√17=3/√17=3√17/17

4、化简cos[(3k+1)/3 *π + α]+cos[(3k-1)/3 * π – α] k∈Z
解：cos[(3k+1)/3 *π + α]+cos[(3k-1)/3 * π – α]
= cos(kπ+π/3 + α)+cos(kπ-π/3 - α)
=cos(kπ)cos(π/3 + α)-sin(kπ)sin(π/3 + α)+cos(kπ)cos(π/3 + α)+sin(kπ)sin(π/3 + α)
=2 cos(kπ)cos(π/3 + α)
当k=2n,n∈Z时，cos[(3k+1)/3 *π + α]+cos[(3k-1)/3 * π – α] =2 cos(kπ)cos(π/3 + α)
=2cos(π/3 + α)=cosα-√3sinα
当k=2n+1,n∈Z时，cos[(3k+1)/3 *π + α]+cos[(3k-1)/3 * π – α] =2 cos(kπ)cos(π/3 + α)
=-2cos(π/3 + α)=-cosα+√3sinα

1,
sinx/(1-cos)*[(tanx-sinx)/(tanx+sinx)]

= sinx/(1-cos)*{tanx(1-cosx)/[tanx(1+cosx)]}

= sinx/(1+cosx)

2,
记 g(x) = |x + 1| - |x - 10|
x <= -1时，g(x) = -x-1 -10+x = -11
-1 < x <= 10时，g(x) = x+1-10+x = 2x-9
10 < x时，g(x) = x + 1 - x + 10 = 11.

log(a)x = 2 + sinβ = lnx/lna.
lnx = [2+sinβ]lna
x = e^[(2+sinβ)lna]

1 <= 2+sinβ <= 3.

当0 < a < 1时，
lna < 0, 0 > lna >= (2+sinβ)lna >= 3lna .

1 = e^0 > e^(lna) = a >= x = e^[(2+sinβ)lna] >= e^(3lna) = a^3 > 0
g(a) = 2a - 9 >= g(x) >= g(a^3) = 2a^3 - 9

当1 < a <= (10)^(1/3)时，
lna > 0, 0 < lna <= (2+sinβ)lna <= 3lna .

1 = e^0 < e^(lna) = a <= x = e^[(2+sinβ)lna] <= e^(3lna) = a^3 <= 10
g(a) = 2a - 9 <= g(x) <= g(a^3) = 2a^3 - 9

当 (10)^(1/3) < a <= 10时，
lna > 0, 0 < lna <= (2+sinβ)lna <= 3lna .
a^3 > 10.

1 = e^0 < e^(lna) = a <= x = e^[(2+sinβ)lna] <= e^(3lna) = a^3
g(a) = 2a - 9 <= g(x) <= g(a^3) = 11

当 a > 10时，
lna > 0, 0 < lna <= (2+sinβ)lna <= 3lna .

10 < e^(lna) = a <= x = e^[(2+sinβ)lna] <= e^(3lna) = a^3
g(x) = 11.

综合，有
当0 < a < 1时，|x+1|-|x-10|的值域为区间 [2a^3 - 9, 2a - 9].

当1 < a <= (10)^(1/3)时，|x+1|-|x-10|的值域为区间 [2a - 9, 2a^3 - 9].

当 (10)^(1/3) < a <= 10时，|x+1|-|x-10|的值域为区间 [2a - 9, 11].

当 a > 10时，|x+1|-|x-10|的值域为区间 [11, 11].

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f(x) = -x^2 + 2x + 3,
f'(x) = -2x + 2 = 2(1-x).
0<=x<=1时，f'(x)>=0,f(x)单调递增，f(0)=3<=f(x)<=f(1)=4.
1f(x)>=f(3)=0.

m=4,n=0.

点P的坐标为（4，-1）。
4^2 + (-1)^2 = 17

sinα = -1/(17)^(1/2)
cosα = 4/(17)^(1/2)

sinα + cosα = 3/(17)^(1/2)

3,
cos(3k+1/3 *π + α）+ cos（3k-1/3 * π - α) = 2cos[3k]cos[1/3 *π + α]

= 2cos[3k]{cosα*(1/2) - sinα*(3^(1/2)/2)}

= cos[3k]{cosα - 3^(1/2)sinα}

1 tanx=sinx/cosx 带入到tanx-sinx/tanx+sinx
可得tanx-sinx/tanx+sinx =(1-cosx)/(1+cosx)

剩下就简单了 可是没有看明白你写的原题(有歧义)

2 2+sinβ 大于1小于3 所以x大于10小于1000(应该是lg吧)
值域是-979到11

0≤x≤3 f(x)单调增 n=3(x=0) m=18(x=3)
P(18,2) sinα+cosα可求