17题三角函数题求三角形面积那个

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 22:10:29
17题三角函数题求三角形面积那个
xWnG~;ŋN߀T3ĥ$ԫ@ 5?DiUQPUAJ ")y!WIe՛739sV4wn\ۻ˭ѭ߁=\Jpg?rEsg>V^|s^O,,/w.,-/ZuOf|B'UK '< O9Asv1VrMhUKV$D@RKr,DY)%E4E6T(͏΄2WX4#TRDP"n X \.f3 [oqDb '+,} y*iԘMܮ9DpJmN\/6jyckzo|(-%܋Q{v6I dm}xY&6J◛}]blb|y-amD6'7R s4DO%L lv׿('^f{N*mV+ ^1Ƶk[ï|k{UUrcjX{`kmyw`}H#=}3jA>}ƅGon^;ǯϸ|ŝ`X9~NX1&:7|

17题三角函数题求三角形面积那个
17题三角函数题

求三角形面积那个

17题三角函数题求三角形面积那个
sin(2A-π/6)+2cos²A-1=1/2;
sin(2A)cos(π/6)-sin(π/6)cos(2A)+cos(2A)=1/2;
(√3/2)sin(2A)-(1/2)cos(2A)+cos(2A)=1/2;
(√3/2)sin(2A)+(1/2)cos(2A)=1/2;
sin(2A+π/6)=1/2;
2A+π/6=π/6+2kπ或5π/6+2kπ;
A=kπ(舍去)或π/3+kπ;
cosA=(b²+c²-a²)/(2bc)=1/2;
b²+c²-1=bc;
(b+c)²-2bc-1=bc;
3bc=4-1=3;
bc=1;
三角形的面积=(1/2)×bc×sinA=(1/2)×1×(√3/2)=√3/4;
如果本题有什么不明白可以追问,

求得A=60
则由余弦定理2bccosA=b^2+c^2-a^2得bc=1
面积S=1/2bcsinA=√3/4

∵sin(2A-π/6)+2cos²A-1=1/2
∴sin2A×√3/2-cos2A×1/2+cos2A=1/2
sin2A×√3/2+cos2A×1/2=1/2
∴sin(2A+π/6)=1/2
∴2A+π/6=π/6或者5π/6
∴A=0(舍去)或者π/3
∵a²=b²+c²-2bccosA
∴1...

全部展开

∵sin(2A-π/6)+2cos²A-1=1/2
∴sin2A×√3/2-cos2A×1/2+cos2A=1/2
sin2A×√3/2+cos2A×1/2=1/2
∴sin(2A+π/6)=1/2
∴2A+π/6=π/6或者5π/6
∴A=0(舍去)或者π/3
∵a²=b²+c²-2bccosA
∴1=b²+c²-bc=(b+c)²-3bc=4-3bc
∴bc=1
∴S⊿ABC=1/2bcsinA=1/2×1×√3/2=√3/4

收起

你做的对的呀!

A+B+C = 180
a = 1
a/sinA = b/sinB = c/SinC
sin(2A-pi/6) + 2cos^2 A - 1 = sin2A sqrt(3)/2 - cos2A 1/2 + cos2A = sin(2A+pi/6) = 1/2
2A + pi/6 = 5pi/6
A = pi/3
a/sinA = 2/sqrt(3)<...

全部展开

A+B+C = 180
a = 1
a/sinA = b/sinB = c/SinC
sin(2A-pi/6) + 2cos^2 A - 1 = sin2A sqrt(3)/2 - cos2A 1/2 + cos2A = sin(2A+pi/6) = 1/2
2A + pi/6 = 5pi/6
A = pi/3
a/sinA = 2/sqrt(3)
b = 2sinB/sqrt(3)
c = 2sinC/sqrt(3) = 2sin(A+B)/sqrt(3)
b+c = 2
sinB+sin(A+B) = sqrt(3)
sinB + sinA cosB + cosA sinB = sqrt(3)/2 cosB + 3/2 sinB = sqrt(3)
3 sinB + sqrt(3) cosB = 2sqrt(3)
sin(B + pi/6) = 1
B = pi/3
A=B=pi/3
a=1=b=c
面积=sqrt(3)/4

收起