若9 cos2a=-根号3 2,则a=5派 k派是他的什么条件

两边平方sin²A+cos²A+2sinAcosA=21+sin2A=2sin2A=1因为sin²2A+cos²2A=1所以cos2A=0再问：为什么sin平方

Bsin^4a+cos^4a=(sin^2a+cos^2a)^2-2(sina*cosa)^2=1-1/2*sin^22a而因为cos2a=根2/3所以cos^22a=2/9sin^2a=7/9代入得

答案是二分之根号二,sin2a=2sinacosa,cos2a=cosa平方-sina平方

∵sina+cosa=√3/3∴(sina+cosa)²=1/3即：sin²a+2sinacosa+cos²a=1/3而sin²a+cos²a=1∴2

(sina+cosa)^2=1\3所以sin2a=2\3.所以cos2a=-根号5\3

sinA/2+cosA/2=2根号3/3两边平方sin²A/2+cos²A/2+2sinA/2*cosA/2=4/31+sin[2*(A/2)]=4/3sinA=1/3cos2A=

∵sina+cosa=√3/3∴(sina+cosa)²=1/3即：sin²a+2sinacosa+cos²a=1/3而sin²a+cos²a=1∴2

sin(a-π/4)/cos2a=[(√2/2)(sina-cosa)]/(cos²a-sin²a)={(√2/2)(sina-cosa)}/{(cosa+sina)(cosa-s

cos2a=(cosa)^2-(sina)^2sin(a-45°)=√2/2(sina-cosa)∴cos2a/sin(a-45°)=-√2(sina+cosa)=-√2/2∴cosa+sina=1/

因为sina^2+cosa^2=1所以cosa^2=8/9cos2a=cosa^2-sina^2=8/9-1/9=7/9

sina+cosa=(1-√3)\2sin^2a+cos^2a=1(sina+cosa)^2=sin^2a+2sinacosa+cos^2a2sinacosa=sin2a=[(1-√3)\2]^2-1

即(cos²a-sin²a)/(sinacosπ/4-cosxsinπ/4)=(cosa+sina)(cosa-sina)/[√2/2(sina-cosx)=-(cosa+sina

cos2a=根号2/3sin2a=±根号5/3sin4a+cos4a=2cos²2a-1+2sin2acos2a=-1/9±根号10/9

cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4))/(2sin(a-π/4)cos(a-π/4))cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4))/sin

sinα+cosα=√3/3(sinα+cosα)²=1/3sin²α+2sinαcosα+cos²α=1/31+sin2α=1/3sin2α=-2/3＜0cos2α=√

∵tan(a+π/4)=3+2√2==>(1+tana)/(1-tana)=3+2√2(应用正切和角公式展开)==>2/(1-tana)=4+2√2(两端同时加1)==>1-tana=1/(2+√2)

分子cos2a=cos^2a-sin^2=(cosa+sina)(cosa-sina),分母sin(a-Pi/4)=(sina-cosa)*（根号2）/2所以由题直接得,答案=1/2

cos(π/4-a)=2√2·cos2acos(π/4)cosa+sin(π/4)sina=2√2(cos²a-sin²a)(√2/2)(cosa+sina)=2√2·(cosa+

cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4))/(2sin(a-π/4)cos(a-π/4))cos2a/sin(a-π/4)=(2cos2a*cos(a-π/4))/sin