Question

在三角形ABC中,化简sin^2A-(sin^2B+sin^2C-2sinBsinCcosA)的结果

^2是平方的意思

Answer 1

由正弦定理有 a=2RsinA,b=2RsinB,c=2RsinC
所以 sin^2A-(sin^2B+sin^2C-2sinBsinCcosA)
=1/（2R）（a^2-b^2-c^2+2bc cosA)
由 余弦定理2bc cosA=b^2+c^2-a^2
所以 sin^2A-(sin^2B+sin^2C-2sinBsinCcosA)=0

Answer 2

三角形面积=1/2ABsinC=1/2BCsinA=1/2ACsinB(A B C为三角形边长)
∴A:sinA=B:sinB=C:sinC=k
有余弦定理：A^2=B^2+C^2-2BCcosA
∴(ksinA)^2=(ksinB)^2+(ksinC)^2-2ksinBksinCcosA
即sin^2A=sin^2B+sin^2C-2sinBsinCcosA
所以sin^2A-(sin^2B+sin^2C-2sinBsinCcosA)=0

Answer 3

解：A=π-(B+C),sinA=sin[π-(B+C)]=sin(B+C),cosA=cos[π-(B+C)]=-cos(B+C)
所以sin^2A=sin^2(B+C)=[sinBcosC+cosBsinC]^2
=(sinBcosC)^2+2sinBcosCcosBsinC+(cosBsinC)^2
cosA=-cos(B+C)=-(cosBcosC-sinBsinC)
2sinBsinCcosA=2sinBsinC[-(cosBcosC-sinBsinC)]=2(sinBsinC)^2-2sinBsinCcosBcosC
所以sin^2A-(sin^2B+sin^2C-2sinBsinCcosA)
=(sinBcosC)^2+2sinBcosCcosBsinC+(cosBsinC)^2-[sin^2B+sin^2C-(2(sinBsinC)^2-2sinBsinCcosBcosC)]
=(sinBcosC)^2+2sinBcosCcosBsinC+(cosBsinC)^2-sin^2B-sin^2C+2(sinBsinC)^2-2sinBsinCcosBcosC
=(sinBcosC)^2+(cosBsinC)^2-sin^2B-sin^2C+2(sinBsinC)^2
=sin^2B(cos^2C-1)+sin^2C(cos^2B-1)+2(sinBsinC)^2
=-sin^2Bsin^2C-sin^2Csin^2B+2(sinBsinC)^2
=0

Answer 4

三角形的三个角满足：A+B+C=π. 首先设法将式子中的A消掉，自然要将A=π-B-C 代人原式。
把 sinA=sin(π-B-C)=sin(B+C), cosA=cos(π-B-C)= -cos(B+C), 代人原式得：
sin^2(B+C)－(sin^2B+sin^2C+2sinBsinCcos(B+C))=
[sinBcosC+cosBsinC]^2－sin^2B－sin^C－2sinBsinC[cosBcosC－sinBsinC]=
sin^2Bcos^2C+2sinBsinCcosBcosC+cos^2Bsin^2C－sin^2B－sin^C－2sinBsinCcosBcosC+2sin^2Bsin^2C=
sin^2Bcos^2C+cos^2Bsin^2C－sin^2B－sin^C+2sin^2Bsin^2C=
-sin^2Csin^2B－sin^2Csin^2B+2sin^2Bsin^2C=0.
故原式化简得结果是0.