高中数学三角函数题，急，谢谢了！ 高中数学三角函数题？
\u9ad8\u4e2d\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u9898\uff1fS\u25b3ABC=1/2acSinB\u3002(\u65e0\u8bba\u521d\u9ad8\u4e2d\uff0c\u6c42\u4e09\u89d2\u5f62\u9762\u79ef\u6700\u5927\u503c\u95ee\u9898\uff0c\u7b2c\u4e00\u4e2a\u8981\u60f3\u5230\u7684\u5c31\u662f\u516c\u5f0f\uff01)
ac\u5df2\u7ecf\u77e5\u9053\u4e86\uff0c\u4ee3\u5165\u516c\u5f0f\u4e2d\u3002
\u8fdb\u4e00\u6b65\u5206\u6790:\u8981\u4f7fS\u6700\u5927\uff0c\u90a3\u4e48SinB\u5c31\u8981\u6700\u5927\u3002
B\u53d6\u503c\u8303\u56f4\u5728(0\uff0c90\u00b0]\uff0c\u6240\u4ee5\u5f53B=90\u00b0\u65f6\uff0cSinB\u6700\u5927\uff0c\u5373S\u25b3ABC\u6700\u5927\u3002
cosA+2cos(B+C)/2
\u56e0\u4e3aA,B,C\u4e3a\u4e09\u89d2\u5f62\u5185\u89d2,\u6240\u4ee5A+B+C=180'
\u6240\u4ee5:cosA+2cos(B+C)/2=cosA+2sinA/2=1-2sin^2(A/2)+2sin(A/2)
\u8bbesin(A/2)\u4e3ax,\u4e140<=x<=1
\u5373\u8f6c\u5316\u4e3a\u4e8c\u6b21\u51fd\u6570
y=1-2x^2+2x
x=-b/2a=1/2(\u7b26\u5408x\u8303\u56f4)\u65f6,y\u53d6\u6700\u5927
y=1.5,\u6b64\u65f6A\u4e3a120'
\u6b64\u9898\u5173\u952e\u662f\u4e09\u89d2\u51fd\u6570\u4e2d\u7684"\u5f02\u540d\u5316\u540c\u540d"
图
A+3C=(π-B-C)+3C=π→2C=B
sinB/sinC=sin(2C)/sinC=2sinCcosC/sinC=2/√3→cosC=⅓√3
cosC=⅓√3→sinC=√(1-⅓)=√⅔→sinB=2sinCcosC=2√⅔·⅓=⅔√2
b/c=2/√3→c=4.5
S=½bcsinA=½·3√3·4.5sin(π-B-C)=6.75√3sin(B+C)
sin(B+C)=sinBcosC+sinCcosB=⅔√2·⅓√3+√⅔·√(1-8/9)
=⅓√6
∴S=6.75√3·⅓√6=6.75√2=
(1)
A+3C=A+B+C
B=2C
由正弦定理得b/sinB=c/sinC
sinC=csinB/b=2sinCcosC/(b/c)=2sinCcosC/(2√3/3)
cosC=√3/3
(2)
sinC=√(1-cos²C)=√[1-(√3/3)²=√6/3
由正弦定理得
sinB=bsinC/c=(b/c)sinC=(2√3/3)(√6/3)=2√2/3
(3)
cosB=cos2C=1-2sin²C=1-2(√6/3)²=-⅓
sinA=sin(B+C)
=sinBcosC+cosBsinC
=(2√2/3)(√3/3)+(-⅓)(√6/3)
=√6/9
b=3√3
c=b/(2√3/3)=(3√3)/(2√3/3)=9/2
S△ABC=½bcsinA
=½·3√3·(9/2)·(√6/9)
=9√2/4
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