已知f(x)=sin(o+x)+sin(o-x)-2sino,o∈(0,3π/2),其中tano=3若对任意X∈R,都有f(x)≥0成立，求cos（o-π/4
\u5df2\u77e5tan2\u03b8=-3/4\u4e14\u03b8\u2208\uff080,3\u03c0/2\uff09\uff0c\u53c8f\uff08x\uff09=sin\uff08\u03b8+x\uff09+sin\uff08\u03b8-x\uff09-2sin\u03b8\u22650\u5bf9\u4e8ex\u2208R\u6052\u6210\u7acb\uff0c\u6c42cos\u56e0\u4e3a f(x)=sin(\u03b8+x)+sin(\u03b8-x)-2sin\u03b8\u22650 \u6052\u6210\u7acb\uff0c\u6240\u4ee5 2sin\u03b8cosx-2sin\u03b8\u22650\uff0c\u6545 sin\u03b8\u22640\uff1b
\u518d\u7531\u6761\u4ef6 \u03b8\u2208(0,3\u03c0/2) \u53ef\u5f97\uff0c\u03c0<\u03b8<3\u03c0/2\uff0c\u2234 2\u03c0<2\u03b8<3\u03c0\uff0c cos\u03b8<0\uff0ccos2\u03b8<0\uff1b
cos2\u03b8=-\u221a[1/(1+tan²2\u03b8)]=-\u221a[1/(1 +9/16)]=-4/5\uff1b
cos\u03b8=-\u221a[(1+cos2\u03b8)/2]=-\u221a[(1 -4/5)/2]=-\u221a10/10\uff1b
f(x) = cos(x+\u03c0/6)-sin(x-2\u03c0/3)+sinx+a
= cos(x+\u03c0/6)-{-sin(\u03c0+x-2\u03c0/3)} + sinx + a
= cos(x+\u03c0/6)+sin(x+\u03c0/3)} + sinx + a
= cos(x+\u03c0/6)+cos{\u03c0/2-(x+\u03c0/3)} + sinx + a
= cos(x+\u03c0/6)+cos{\u03c0/6-x} + sinx + a
= cos(x+\u03c0/6)+cos{x-\u03c0/6} + sinx + a
= cosxcos\u03c0/6-sinxsin\u03c0/6 + cosxcos\u03c0/6+sinxsin\u03c0/6 + sinx + a
= 2cosxcos\u03c0/6 + sinx + a
= 2(cosxcos\u03c0/6 + sinxsin\u03c0/6) + a
= 2cos(x-\u03c0/6)+a
-2+a\u22642cos(x-\u03c0/6)+a\u22642+a
2+a=1
a=-1
f(x)=sin(o+x)+sin(o-x)-2sino
=2sino*cosx-2sino
=2sino(cosx-1)≥0
∵cosx-1恒≤0 ∴sino＜0
o∈(π,3π/2)
由tano=3 容易得到 sin0=(-3√10)/10 coso=(-√10)/10
由两角差的余弦公式：
cos（o-π/4）=(√2/2)*(sino+coso)=（-2√5)/5
f(x)=sin(o+x)+sin(o-x)-2sino,o∈(0,3π/2), f(x)≥0可知sino=《0 ,0在第三象限
tano=3，得coso=-1/根号10.sino=-3/根号10
cos（o-π/4）=-2根号5/5
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