cos2a=3 5,sinb=10 10,若a-b=2-1求a,b,c

因为sinA+cosA=2sina两边平方得sin^2A+cos^2A+2*sinA*cosA=4*sin^2a则2*sinA*cosA=4*sin^2a-1因为sinA*cosA=(sinb)^2则

cos2a=1-2(sina)^2=1-2/9=7/9

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

(sinC+cosC)/2=sinA;sinB/sinC=cosC/sinB;顺序分析法：2cos2A=cos2B；2（1-2sinA^2）=1-2sinB^22[1-2((sinC+cosC)/2)

这是一道中等难度的题目,看我解答啊（1）首先2A的余弦等于0.6,根据倍角公式,A的余弦等于2倍根号5除以5,再根据和角公式可以算出A加B的余弦等于根号2除以2,那么A加B角不就等于45度（2）根据正

∵sina=-3/5,a∈(∏,3∏/2),∴cosa=-√(1-sin²a)=-4/5∵sinb=12/13,b∈(∏/2,∏),∴cosb=-√(1-sin²b)=-5/13∴

sinA=1/3则cos2A=1-2sin²A=1-2*(1/3)²=1-2/9=7/9

(sinA+sinB+sinC)²=sin²A+sin²B+sin²C+2(sinAsinB+sinAsinC+sinBsinC)(cosA+cosB+cosC

cosa+cosb=0……(1)sina+sinb=1……(2)(1)^2+(2)^2:1+2(cosacosb+sinasinb)+1=1--->cos(a-b)=-1/2.(1)--->2sin[

cos2A+cos2B=2cos²A-1+2cos²B-1∵cosA+cosB=0∴cosA=-cosB∴=4cos²B-2∵sin²B+cos²B=

sin2a+sina/cos2a+cosa+1=(2sinacosa+sina)/(2cos²a+cosa)=sina(2cosa+1)/[cosa(2cosa+1)]=sina/cosa=

证明:1-cos(2A)=2*[(sinA)^2]1+cos(2A)=2*[(cosA)^2]sin(2A)=2sinA*cosA==>(1+sin2A-cos2A)/(1+sin2A+cos2A)=

证：∵（sin2a-cos2a)^2=sin²2a+cos²2a-2sin2acos2a而sin²2a+cos²2a=1,2sin2acos2a=sin4a,∴

分子=sin²a+cos²a+2siacosa-(cos²a-sin²a)=(sina+cosa)²-(cosa+sina)(cosa-sina)=(

：①∵sinB=1/√10,B为锐角,∴cosB=3/√10.∵cos2A=3/5,A为锐角,∴2sin²A=1-cos2A=2/5∴sinA=1/√5,cosA=2/√5.∵sin（A+B

sina+sinb=0则sina=-sinb即a=-b,则有2cosa=2/3,cosa=1/3.cos2a+cos2b=4cosa方-2=-14/9

(sin2a-cos2a)^2=sin2a^2+cos2a^2-2*sin2a*cos2a=1-2*sin2a*cos2a=1-sin4a因为sin2a^2+cos2a^2=12*sin2a*cos2

cos2a=2cosa的平方-1=3/5tan[a+b]=sin[a+b]/cos[a+b]sin(a+b)=sinacosb+cosasinbcos(a+b)=cosacosb-sinasinbsi

sin2a=2sina*cosacos2a=cos²a-sin²a=2cos²a-1=1-2sin²a所以(1+cos2a)/sin2a-sin2a/(1-co