一道高中数学三角函数题求解

2024-11-06 15:32:32
推荐回答(2个)
回答(1):

要用到的一个公式:cos2a=cos²a-sin²a=1-sin²a-sin²a=1-2sin²a
所以
2sin²a=1-cos2a
解答:
b²+c²=a²+bc
则b²+c²-a²=bc
根据余弦定理得到:cosA=(b²+c²-a²)/(2bc)=bc/(2bc)=1/2
而0<A<180°

所以A=60°
由于A+B+C=180°
所以B+C=120°
B=120°-C
而2sin²B/2+2sin²C/2=1
即1-cosB+1-cosC=1
所以cosB+cosC=1

cos(120°-C)+cosC=1
展开得到:cos120°cosC+sin120°sinC+cosC=1
所以
-1/2cosC+√3/2sinC+cosC=1
得到:1/2cosC+√3/2sinC+cosC=1
即cosC*cos60°+sin60°*sinC=1
得到:cos(C-60°)=1
所以C-60°=0

即C=60°

于是B=120°-60°=60°
综合得到:A=B=C=60°
所以△ABC为等边三角形。
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回答(2):

因为f(0)=2,f(П/3)=1/2+√3/2
;
所以f(0)=2a=2,f(П/3)=a/2+b*√3/4=1/2+√3/2
;
所以a=1,b=2;
所以f(X)=2cos^X+2sinXcosX;
所以f(X)=cos2X+sin2X+1=(√2)*sin(2X+П/4)+1;
所以max=1+根号2,min=-1-根号2;
因为(√2)*sin(2a+П/4)+1=0;
所以sin(2a+П/4)=-1/√2;
所以2a+П/4=-П/2+П/4+2П或=-П/2-П/4+2П;
所以a=П/2或3П/4

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