cos2c等于

sina(sinb+cosb)-sinc=0sinasinb+sinacosb-sin(a+b)=0sinasinb+sinacosb-(sinacosb+cosasinb)=0sinasinb-co

sinA(sinB+cosB)-sinC=0-----sinB(sinA-cosA)=0sinA=cosA,A=45sinB-cos2C=sinB-cos(2B+90)=sinB-sin2B=sinB

cos2A+cos2B=2cos(A+B)cos(A-B)1+cos2C=2(cosC)^2cos(A+B)=-cosC-cosCcos(A-B)=(cosC)^2所以cosC=0或-cos(A-B)

(1)cos2C=1-2(sinC)^2=-1/9sinC=√5/3(2)2sinC等于根五倍的sinA,得2c=√5a,c=√5,cos=2/34+b^2-2*2bcosC=5,b=3b=3,c=√

∵cos2c=1-2sin^2C∴sinC=√10/4CosC=√(1-10/16)=√6/4CosA=√(1-10/64)=3√6/8b=AD+CD=4*3√6/8+2*√6/4=2√6根据正弦定理

(1)因为cos2C=cos（C+C）=（cosC）平方-（sinC）平方=-3/4（cosC）平方+（sinC）平方=1得（sinC）平方=7/8所以,sinC=根号14/4（2）

cos2C=1-2sin^2C=－3／4则sin^2C=(1+3/4)/2=7/8sinC=√14/4当c＝2a且b=3√7时,由cosC=√(1-sin^2C)=√2/4所以在三角形中,由余弦定理得

cos2C=-1/4所以cos2c=1-2(sinc)^2=-1/4可得sinc=+—根号10/4又因为角c为三角形内角所以正弦值是正数所以sinc=根号10/4

1－cos2C=2(sinC)^2三角形中sinC>0则sinC=2b/a=2sinB/sinA=2sin(A+C)/sinA1/2sinA*sinC=sinA*cosC+sinC*cosA同除以si

△ABC中，tanA，tanB是3x+8x-1=0的两个实数根，可得tanA+tanB=-83，tanAtanB=-13，所以tan（A+B）=tanA+tanB1-tanAtanB=-8343=-2

sinC=sin(π-(A+B))=sin(A+B)cos2C=cos2(π-(A+B))=cos2(A+B)∴sinA(sinB+cosB)-sin(A+B)=0sinAsinB+sinAcosB)

cos2A+cos2B+cos2Ccos2A+cos2B+cos2C=(cos2A+cos2B)+(cos2B+cos2C)+(cos2A+cos2C).用和差化积公式cos(a)+cos(b)=2c

∵A，B，C成等差数列，∴2B=A+C，又A+B+C=π，∴B=60°，即A+C=120°，cos2A+cos2C=1+cos2A2+1+cos2c2=1+cos2A+cos2C2=1+cos（A+C

由正弦定理得a/sinA=b/sinB,因为acosA=bcosB,所以sinAcosB-cosAsinB=sin（A-B）=0,所以∠A=∠B.cos2A+cos2B-cos2C=2cos2A-co

cos2A+cos2B=2cos(A+B)cos(A-B)1+cos2C=2(cosC)^2cos(A+B)=-cosC-cosCcos(A-B)=(cosC)^2所以cosC=0或-cos(A-B)

超简单~的思路一般都是联系条件（三角形）那么就把“大角化小角,一步一步慢慢走咯（柴）”.1-2sin^2A+1-2sin^2B-1+2sin^2C=1得sin^2A+sin^2B=sin^2C在用上正

sin²2C+sin2C×sinC+cos2C=1,4sin²C*cos²C+2sin²CcosC+1-2sin²C=1,2cos²C+co

这属于多变量的极值问题,可以采取所谓的“冻结变量法”.显然A,B,C三个角中至少有两个锐角,不妨假设C为锐角,固定角C不变,由和差化积公式：cos2A+cos2B=2cos(A+B)cos(A-B)=

sin(A+B)=cos2Csin(180-C)=1-2sin²CsinC=1-2sin²C2sin²C+sinC-1=0(sinC+1)(2sinC-1)=0C是内角则

根据正弦定理,(a+b)/a=sinB/(sinB-sinA)=(sinA+sinB)/sinA∴sinA·sinB=(sinB+sinA)(sinB-sinA)=2sin[(B+A)/2]·cos[