cos2c/2等于?

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

cos2C+2cos(180-C)=-3/2cos2C-2cosC=-3/22cos^2C-2cosC+1/2=0(^2表示平方)(2cosC-1)*2=0cosC=1/2C为60度c^2=a^2+b

cos2C＝-1/4=1-2sin^2CsinC=根号10/42sinA＝sinC2a=cc=4cosC=根号6/4sinA=根号10/8cosA=根号54/8cosB=-cos(A+C)=1/4b^

(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

第一问,cos2c=2coscsinc=-1/4sinc平方+cosc平方=1联立求解再答：第二问因为2sinA=sinC，所以2a=c,c=4,然后再使用余弦定理cosc展开，即得b再答：明白没再问

(1)cos2C=1－c^2/b^2==>1－2(sinC)^2＝1－(sinC/sinB)^2==>sinB＝√2/2==>B＝45°(锐角三角形)(2)cosA＝3/5==>sinA＝4/5根据正

在锐角三角形ABC中,cos2C+2=2sin^2C=1-cos2C,∴cos2C=-1/2,2C=2π/3,C=π/3.（1）B=5π/12,∴A=π/4,由正弦定理,a=csinA/sinC=√1

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

2sina=sinb+cosb4sin²a=sin²b+cos²b+2sinbcosb=1+2sinbcosbsin²c=sinbcosb所以4sin²

当a=2,2sinA=sinC时,由正弦定理asinA=csinC,得：c=4由cos2C=2cos2C-1=-14,及0＜C＜π得cosC=±64由余弦定理c2=a2+b2-2abcosC,得b2±

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

∵ab≤(a+b)^2/4=25/4ab最大值25/4＜9CosC=(a^2+b^2-c^2)/2ab=[(a+b)^2-c^2-2ab]/2ab=(25-7-2ab)/2ab=(18-2ab)2ab

(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

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

（1）由2sin2C-2cos2C=1有：cos2C＝cos2C−sin2C＝−12（3分）（也可将1化为1=sin2C+cos2C，转化为tanC求解C）∵C∈（0，π2）∴2C=2π3，从而有：C

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