cos2c=-4分之1

1.cosC=2cosA1:AB=1:根号52.cosA=3/(10*根号5)将原式化简出并带入COSA和sinA即可

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)

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

△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

4cos²C/2-cos2C=7/22(1+cosC)-2cos²C+1=7/24cos²C-4cosC+1=0(2cosC-1)²=0cosC=1/2C=60

∵cos2c=1-sin^2 C ∴sin C=√10 /42）Sin A=1/2 sin C = √10&n

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再问：其实这道题我已经做出来了、不过还

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

4sin^2(A+B)-cos2C=2[1-cos(A+B)]-(2cos^2C-1)=7/2,4cos^2C-4cosC+1=0cosC=1/2,C=60度a+b=5,a^2+2ab+b^2=25,

单独求第二问,根据正弦定理,得a/sinA=c/sinC,∴c=asinC/sinA=4,cos2C=2cos²-1,∴cosC=√6/4或-√6/4,cosC=(a²+b

(1)根据cos2C=1-2sin²C,得sinC=±√10/4,又C为三角形内角,得sinC=√10/4(2)根据正弦定理,a=c/2,sinA=√10/8,c=4,b=2√6

cos2C=2cos^2C-1=-3/4cos^2C=1/8cosC=√2/4余弦定理√2/4=(b^2+a^2-c^2)/2abc=2a√2/4=(63+a^2-4a^2)/6√7a2a^2+√14

(1)cos2C=1-2sin²c=-1/92sin²c=10/9sin²c=5/9sinC>0sinC=√5/3cosC=2/3(2)a=2sinC/sinA=c/a=

4sin²（A+B/2）-cos2C=2-2cos（A+B）-(2cos²C-1）=3+2cosC-2cos²C=7/2所以cosC=1/2,所以∠C=60°cosC=a

/>4sin²[(A+B)/2-cos2C=7/2,2(1-cos(A+B))-(2cos²C-1)=7/2,2(1+cosC)-(2cos²C-1)=7/2,2+2co

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[

a=1,c=根号7,则解法如下：∵4sin^2（A+B）/2）-cos2C=7/24sin^2[(180°-C)/2]-cos2C=7/24sin^2(90°-C/2)-cos2C=7/24cos^2