高一数学：三角函数问题？ 高一数学：三角函数问题？
\u9ad8\u4e00\u6570\u5b66\uff1a\u4e09\u89d2\u51fd\u6570\u95ee\u9898\uff1f\u7528\u53cd\u4e09\u89d2\u51fd\u6570\u8868\u793a\uff0c\u4e0d\u8fc7\u8981\u5173\u4e8e\u03c0/2\u5bf9\u79f0\uff0c\u8be6\u7ec6\u8fc7\u7a0b\u8bf7\u89c1\u56fe\u7247
(1) cos\u03b1 = 1/7, \u56e0\u4e3a0\uff1c \u03b1\uff1c\u03c0/2 ,
\u6240\u4ee5sin\u03b1 = \u221a(1-cos²\u03b1) = \u221a[1-(1/7)²] = 4 \u221a 3 / 7
\u6240\u4ee5tan\u03b1 = sin\u03b1 / cos\u03b1 = 4 \u221a 3
+++++++++++++++++
tan 2\u03b1=((2tan\u03b1)/(1-tan^2 \u03b1) )=8\u221a 3/(1-48)=-8\u221a 3/47;
(2) cos\uff08\u03b1-\u03b2\uff09=13/14, \u56e0\u4e3a -\u03c0/2 \uff1c \u03b1 -\u03b2 \uff1c\u03c0/2,
\u6240\u4ee5sin\uff08\u03b1-\u03b2\uff09 = \u221a [1-(cos²(\u03b1-\u03b2)] = \u221a [1-(13/14)²] = 3\u221a3 /14
\u6839\u636e\u4e24\u89d2\u5dee\u7684\u4f59\u5f26\u516c\u5f0f:
cos[\u03b1 - (\u03b1-\u03b2)] = cos\u03b1cos(\u03b1-\u03b2) + sin\u03b1sin(\u03b1-\u03b2)
cos\u03b2 = (1/7) * (13/14) + (4 \u221a 3 / 7) * (3\u221a3 /14)
= 1/2
\u6240\u4ee5\u03b2=-pi/3;
\u4e0d\u61c2\u518d\u8ffd\u95ee\uff0c\u6ee1\u610f\u8981\u70b9\u4e2a\u91c7\u7eb3\u5440\u3002
依题意f(x)=0在定义域只有一解，等价于k=g(t)=\sqrt{2}*sin(t+pi/4)+2*sin(2*t)在定义域内只有一解
当t=pi/4时，sin(t+pi/4)和sin(2*t)同时取最大值，g(pi/4)=2+\sqrt{2}
当t=0或t=pi/2（不可取到）时，sin(t+pi/4)和sin(2*t)同时取最小值, g(0)=1
当0<t<pi/4时，g(t)单调递增；当pi/4<t<pi/2时，g(t)单调递减
所以实数k的值可以是2+\sqrt{2}或1，分别对应的唯一零点为t=pi/4, t=0，即x=3*pi/4, x=pi/2
g(x) = |sinx|+|cosx|-4sinxcosx = |sinx|+|cosx|-2sin2x
在区间[π/2，π)上，g(x)于π/2取得最小值仅一次:g(π/2) = 1。所以，k = 1, y = f(x)在区间[π/2，π)上恰好有1个零点.
f(x)=|sinx|+|cosx|-4sinxcosx-k，在区间[π/2，π)上
=sinx-cosx-4sinxcosx-k
=4[sin(x-π/4)-根2/8]^2-2-k-1/8
x=3π/4时，f(x)最大>=0 x=π/2或π时，f(x)最小=<0
1<=k<=2+根2
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