在角ABC中,8sin方
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sinA+cosA=1/5(sinA+cosA)^2=1/251+2sinAcosA=1/25sinAcosA=-12/25∵sinA>0∴cosA<0∴sinA-cosA>0∴sinA-cosA=√
SINA方=SINB方+SINBSINC+SINC方根据正弦定理a/sinA=b/sinB=c/sinC转化a^2=b^2+c^2+bcbc=-(b^2+c^2-a^2)余弦定理cosA=(b^2+c
由正弦定理,a/sinA=b/sinB=c/sinC=2R,得sinA=a/2R,sinB=b/2R,sinC=c/2R从而由sin²A=sin²B+sin²C,得a
a/sinA=2R所以a^2+b^2a^2+b^2所以2abcosC
sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理:cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以:cosC
"AB=(40000-1)2+4002=因为以上规律总结得到((n+1)2-1)2+(2(n+1))2=((n+1)2+1)2其中n=1满足3方+4方=5方令(2(n+1))2=4002((n+1)2
S=1/2absinC且cosC=(a^2+b^2-c^2)/(2ab),由题目知道S=(a^2+b^2-c^2)/4,对比三个公式,可以得出:S=1/2absinC=1/2abcosC,所以sinC
解,8sin²((B+C)/2)-2cos(2A)=78cos²(A/2)-2(2cos²A-1)=74cosA+4-4cos²A+2=7整理,4cos
由正弦定理,a/sinA=b/sinB=c/sinC=2R,得sinA=a/2R,sinB=b/2R,sinC=c/2R从而由sin²A=sin²B+sin²C,得a
1.题目中是(B+C)/2吧?4sin^2(B/2+C/2)-cos2A=2[1-cos(B+C)]-2cos^2A+1=2(1+cosA)-2cos^2A+1=7/2=>cosA=1/2=>A=60
sin方A+sin方B=sin方C根据正弦定理:a/sinA=b/sinB=c/sinC=2Ra^2/(2R)^2+b^2/(2R)^2=c^2/(2R)^2即:a^2+b^2=c^2,符合勾股定理,
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),(sin^A+sin^B)sin(A-B)=(sin^A-sin^B)sin(A+B)sin^A*(sin(A+B)-sin(A
a²≤b²+c²-bcbc≤b²+c²-a²1/2≤(b²+c²-a²)/2bccosa≥1/2a≤60°
a/sinA=b/sinB=c/sinC=2R=2√2=>a=2RsinA,b=2RsinB,c=2RsinC2√2(sin²A-sin²C)=(a-b)sinB=>4R²
【1】sin方a+sin方b+sin方c=sin方a+sin方b+sin方(180-(a+b))=sin方a+sin方b+sin方(a+b)=sin方a+sin方b+(sina*cosb+cosa*s
选A.因为在三角形ABC中,若sinC=sinA+sinB,又因为sinC=sin(180°-A-B)=sin(A+B)=(sinAcosB+sinBcosA)=sinAcosB+2sinAcosAs
余弦定理,a+b-2abcosC=c,而a+b-3/2ab=ccosC=3/4.降次公式得sin(A+B)/2=(1-cos(A+B))/2=(1+cosC)/2=7/8
由正弦定理a/sinA=b/sinB=c/sinC=2R,sin²A+sin²B=sin²C两边同乘以4R²得(2RsinA)²+(2RsinB)
已知sin^2A+sin^2C-sin^2B=sinAsinC由正弦定理知a^2+c^2-b^2=ac∴又由余弦定理知cosB=(a^2+c^2-b^2)/2ac=1/2∴B=60°注意:sina的平