关于高中数学一道三角函数题!
\u4e00\u9053\u9ad8\u4e2d\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u9898\u8bc1\u660e\uff1a\u7531\u9898\u77e5\uff1aC-B=B-A\uff0c\u5373\uff1aA+C=2B\uff0c\u5219A+B+C=3B=180\u00b0\uff0c\u5f97B=60\u00b0\u3002
\u82e5\u25b3ABC\u7684\u4e09\u4e2a\u5185\u89d2A,B,C\u6240\u5bf9\u5e94\u7684\u4e09\u8fb9\u5206\u522b\u4e3a\uff1aa\u3001b\u3001c\uff0c\u7531\u4f59\u5f26\u5b9a\u7406\uff0c\u5f97
b^2=c^2+a^2-2ca*cosB
=c^2+a^2-2ca*cos60\u00b0
=c^2+a^2-2ca*1/2
=c^2+a^2-ca
\u6b32\u8bc1\u7b49\u5f0f\u5de6\u8fb9\uff1a
1/(a+b)+1/(b+c)
=(a+2b+c)/(a+b)(b+c)
=(a+2b+c)/(ab+ac+b^2+bc)=3/(a+b+c)..................\u2460
\u4e8e\u662f\u539f\u9898\u7b49\u4ef7\u4e8e\u8bc1\u660e\u2460\u5f0f\u6210\u7acb\uff0c\u4ea4\u53c9\u76f8\u4e58\u5f97\uff1a
3(ab+ac+b^2+bc)=(a+b+c)(a+2b+c)=(a+b+c)[(a+b+c)+b]
3(ab+ac+b^2+bc)=(a+b+c)^2+b(a+b+c)
3ab+3ac+3b^2+3bc=a^2+b^2+c^2+2ab+2bc+2ca+ba+b^2+bc
\u6574\u7406\uff0c\u5f97
b^2=c^2+a^2-ca\uff0c............................\u2461
\u4e8e\u662f\u8981\u8bc1:1/(a+b)+1/(b+c)=3/(a+b+c)\u6210\u7acb\uff0c\u5c31\u7b49\u4ef7\u8bc1\u660e\u2461\u5f0f\u6210\u7acb\u3002\u800c\u2461\u5f0f\u5df2\u7ecf\u7531\u4f59\u5f26\u5b9a\u7406\u8bc1\u5f97\u3002
\u6240\u4ee5\u7531\u6b64\u5012\u63a8\u5373\u5f97\u3002
sinC=sin(\u03c0-(A+B))=sin(A+B)
cos2C=cos2(\u03c0-(A+B))=cos2(A+B)
\u2234sinA(sinB+cosB)- sin(A+B)=0
sinAsinB+sinAcosB)- sinAcosB-cosAsinB=0
sinB(sinA-cosA)=0,\u53c8sinB\u22600
\u2234sinA=cosA
\u2235A\u2208(0,\u03c0)
\u2234A= \u03c0/4
\u518d\u7531sinB+ cos2(A+B)=0
sinB+cos( \u03c0/2 +2B)=0
sinB-sin2B=0
cosB=- 1/2
\u2234B= \u03c0/3
\u53c8C=\u03c0-(A+B)=\u03c0-( \u03c0/3 + \u03c0/4 )= 5\u03c0/12
\u2234A= \u03c0/4 ,B= \u03c0/3 ,C= 5\u03c0/12 .
(1)f(a)=√2/2sin(a+π/4)=√2/4
得到sin(a+π/4)=1/2
又a属于(0,π) 所以a+π/4属于(π/4,5π/4)
所以a+π/4=5π/6 得到a=7π/12
(2)x属于[-π/4,π] 得到x+π/4属于[0,5π/4] sin(x+π/4)属于[-√2/2,1]
得到f(x)属于[-1/2,√2/2]
所以函数的最大值是√2/2,最小值是-1/2
f(x)=sinx/2×cosx/2+cos^2x/2-1/2
=1/2sinx+1/2(1+cosx)-1/2
=1/2sinx+1/2cosx
=√2/2(√2/2sinx+√2/2cosx)
=√2/2sin(x+π/4)
∵f(a)=√2/4
∴√2/2sin(a+π/4)=√2/4
∴sin(a+π/4)=1/2
∵a+π/4∈(π/4,5π/4)
∴a+π/4=5π/6
∴a=7π/12
(2)
∵x∈(-π/4,π)
∴0<x+π/4<5π/4
∴x+π/4=π/2即x=π/4时,f(x)取得最大值√2/2
因为开区间,f(x)无最小值
值域为(-1/2,√2/2]
f(x)=sinx/2×cosx/2+cos^2x/2-1/2
=1/2sinx+1/2cosx
=√2/2sin(x+π/4)
1)√2/2sin(a+π/4)=√2/4
∵a∈(0,π)
∴a=5π/6-π/4=7π/12
2)x∈(-π/4,π/4)
x+π/4∈(0,π/2)
f(x)∈(0,√2/2)
最大值√2/2,最小值0
f(x)=f(x)=sinx/2×cosx/2+cos^2x/2-1/2
=1/2sinx+1/2(1+cosx)-1/2
=1/2sinx+1/2cosx
=√2/2(√2/2sinx+√2/2cosx)
=√2/2sin(x+π/4)
∵f(a)=√2/4
∴√2/2sin(a+π/4)=√2/4
∴sin(a+π/4)=1/2
∵a+π/4∈(π/4,5π/4)
∴a+π/4=5π/6
∴a=7π/12
f(x)=根号2/2sin(x+派/4),
f(a)=根号2/4
所以sin(x+派/4)=1/2
因为a属于(0,派)
所以a=105度
因为x属于(-派/4,派)
所以x+派/4属于(0,5派/4)
所以f(x)属于(-根号2/2,1)
求最佳!
f(x)=sinx/2+(1+cosx)/2-1/2=√2/2sin(x+π/4),f(a)=√2/4,∴sin(x+π/4)=1/2,又∵a∈(0,π),∴x+π/4=5π/6,即x=7π/12.
x∈(-π/4,π),x+π/4∈(0,5π/4),sin(x+π/4)∈(-√2/2,1],∴f(x)∈(-1/2,√2/2],∴f(x)的最大值是√2/2,无最小值
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