高一数学三角函数题（有关正弦、余弦定理）
\u9ad8\u4e2d\u6570\u5b66\u9898\u6c42\u89e3\uff0c\u4e09\u89d2\u51fd\u6570 \u6b63\u5f26\u4f59\u5f26\u5b9a\u7406\u89e3
\uff08a³+b³-c³)/(a+b-c)=c²
a³+b³-c³=(a+b-c)c²=(a+b)c²-c³
a³+b³=(a+b)c²
(a+b)c²=a³+b³=(a+b)(a²-ab+b²)
c²=a²-ab+b²
a²+b²-c²=ab
(a²+b²-c²)/(2ab)=1/2
\u7ed3\u5408\u4f59\u5f26\u5b9a\u7406\uff1acosC=(a²+b²-c²)/(2ab)=1/2
\u518d\u7ed3\u54080º\uff1cC\uff1c180º\u53ef\u5f97\uff1aC=60º
A+B=120º
3/4=sinAsinB=(1/2)[cos(A-B)-cos(A+B)]
3/2=cos(A-B)-cos120º=cos(A-B)+(1/2)
\u2234cos(A-B)=1
\u7ed3\u5408-120º\uff1cA-B\uff1c120º\u53ef\u77e5\uff1aA=B=60º
\u2234\u8be5\u4e09\u89d2\u5f62\u4e3a\u7b49\u8fb9\u4e09\u89d2\u5f62\u3002
sin²A+cos²A=1
sinA>0
所以sinA=3/5
sinB=√3/2,cosB=1/2
所以sinC=sin(180-A-B)
=sin(A+B)
=sinAcosB+cosAsinB
=(3+4√3)/10
2、
a/sinA=b/sinB
a=bsinA/sinB=6/5
S=1/2absinC=(6√3+24)/25
1、
sin2a=2sinacosa=24/25
sin²a+cos²a=1
所以sin²a+2sinacosa+cos²a=1+24/25
(sina+cosa)²=49/25
a在第三象限，sina<0,cosa<0
sina+cosa=-7/5
sinacosa=12/25
所以sina和cosa是方程x²+7x/5-12/25=0的根
x=-3/5,x=-4/5
5π/4<a<3π/2
所以tana>tan5π/4=1
sina/cosa>1
cosa<0
所以sina<cosa
所以sina=-4/5,cosa=-3/5
所以seca=1/cosa=-5/3
csca=1/sina=-5/4
5π/4<a<3π/2
5π/8<a/2<3π/4
所以cota/2<0
cosa=cos²(a/2)-sin²(a/2)=[cos²(a/2)-sin²(a/2)]/[cos²(a/2)+sin²(a/2)]
上下除以sin²(a/2)
cosa=[cot²(a/2)-1]/[cot²(a/2)+1]=-3/5
cot²(a/2)=1/4
cot(a/2)=-1/2
2、
=-cos(π/7)cos(2π/7)cos(3π/7)
=cos(π/7)cos(2π/7)cos(4π/7)
=2sin(π/7)cos(π/7)cos(2π/7)cos(4π/7)/2sin(π/7)
=sin(2π/7)cos(2π/7)cos(4π/7)/2sin(π/7)
=2sin(2π/7)cos(2π/7)cos(4π/7)/4sin(π/7)
=sin(4π/7)cos(4π/7)/4sin(π/7)
=2sin(4π/7)cos(4π/7)/8sin(π/7)
=sin(8π/7)/8sin(π/7)
=-sin(π/7)/8sin(π/7)
=-1/8
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