The Beatles,since many of you

sinA:sinB:sinC=2:3:4a:b:c=2:3:4(4K)²=(2k)²+（3K)²-2*2k*3k*cosCcosC=-1/4sinC=√15/4

A=B=C=6时0最大,为3/2根号3证明：sinA+sinB+sinc=2sin[(A+B)/2]cos[(A-B)/2]+sinC>=2sin[(A+B)/2]+sinC=2sin(90-C/2)

∵0∴0∴cos(C/2)>sin(C/2).又∵0∴-π∴-π/2∴cos((A-B)/2)>0,∴sin(A)+sin(B)=2sin((A+B)/2)cos((A-B)/2)=2sin((π-C

题目应该是在锐角三角形中.诚如是,则解答如下：先证明sinA+sinB>1+cosC.由A、B是锐角得A-B0,所以sinA+sinB>1+cosC.所以sinA+sinB+sinC>1+cosC+s

t=-50:1:50;y=sin(pi*t/8)./(pi*t/8);xlabel('t','FontSize',20);ylabel('p','

1.假设a/sinA=b/sinB=c/sinC=2R那么sinA=a/2RsinB=b/2RsinC=c/2R因为(sinA)平方=(sinB)平方+sinC(sinB+sinC）所以（a/2R）^

C=180-(A+B)而sin(180-x)=sinx所以sinC=sin[180-(A+B)]=sin(A+B)

就是转置了一下,行向量变成列向量,比如:>>t=0:0.1:10;>>size(sinc(t))ans=1101>>size(sinc(t'))ans=1011>>size(sinc(t.'))ans

证明：设sinA/a=sinB/b=sinC/c=k,则sinA=ak,sinB=bk,sinC=ck,sinA/(sinB+sinC)+sinB/(sinA+sinC)+sinC（sinA+sinB

证:∵△ABC为锐角三角形,∴A+B>90°得A>90°-B∴sinA>sin(90°-B)=cosB,即sinA>cosB,同理可得sinB>cosC,sinC>cosA上面三式相加:sinA+si

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

1.to2.with3.about4.for5.on6.over7.for8.in9.like10.through

1.sinC+cosC化成半角,2sinc/2cosc/2+1-2sinc/2sinc/2原式化为cosC/2-sinC/2=0两边平方,得到1-sinC=0即sinC=12.条件不足,看看题是否写错

ect（x)

可见,你给的代码里,画的是Sa(t).

（1）sinC+cosC=1-sinC/2,移项得sinC-sinC/2=1-cosC由二倍角公式得2sinC/2cosC/2-sinC/2=2(sinC/2)^2因为sinC/2≠0,所以两边消去s

c=40度(~40.48884564)

sinc函数有两个定义,有时区分为归一化sinc函数和非归一化的sinc函数.它们都是正弦函数和单调递减函数1/x的乘积：sinc(x)=sin(pi*x)/(pi*x);归一化rectxsinc函数

∵acosA+bcosB=ccosC∴sinAcosA+sinBcosB=sinCcosC∴sin2A+sin2B=sin2C=sin(2π-2A-2B)=-sin(2A+2B)∴0=sin2A+si