a为钝角 sin(π 4 a)=3 4 求sin(π 4 - a)

我已近毕业好多年了很久没看数学书了用个笨方法解别笑话我哈向量CA=向量OA-向量OC向量CB=向量OB-向量OC向量CA*向量CB=OA*OB-OC(OA+OB)+OC^2=0-2（cosa+sina

答案：A解析：∵(sina+cosa)^2=1+2sinacosa=49/144即：sin2A=49/144-1=-95/144180即A>90故是钝角三角形

/>a是锐角π/2

cos(2α+π/12)=cos(α+π/6+α+π/6-π/4)=cos(α+π/6)cos(α+π/6-π/4)-sin(α+π/6)sin(α+π/6-π/4)=(-2/3)√2cos(α+π/

由sin(α+π/12)=1/3知：cos(α+π/12)=+根号（1-sin^2(α+π/12)）or-根号（1-sin^2(α+π/12)）然后根据公式：cos(α+5π/12)=cos[π/3+

由正弦定理,c/a=sinC/sinA=sin(A+B)/sinA=2,c=4;cos(A+B-C)=cos(π-2C)=-cos(2C)=-2(cosC)^2+1=1/4,cosC=-sqrt(6)

cos(A+B-C)=1/4cos(180°-C-C)=1/4cos2C=-1/42cos^2C-1=-1/4cos^2C=3/8∵C是钝角∴cosC=-√6/4sinC=√(1-cos^2C)=√(

(cosa-sina)^2=1-2sincosa=1-1/4=3/4π/4＜a＜π/2∴cosa

用正弦定理化为a^2>b^2+c^2即b^2+c^2-a^2

sin(45-a)=sin(90-45-a)=sin(90-(45+a))=cos(45+a)又sin(a+45度)＝1/3>0,a+45为钝角所以,原式=-2倍根号2比上3

cosα=-√1-(sinα)^2=-3/5,cosβ=√1-(sinβ)^2=5/13,cos(α-β)=cosαcosβ+sinαsinβ=-3/13+48/65=33/65cos[(α-β)/2

cosa=-3/5,cosb=5/13,cos(a-b)=cosa*cosb+sina*sinb=33/65,再除以2就好啦,不会打阿尔法贝塔就用ab代替了我名字缩写也是zlj呢,真有缘==再问：真的

cosα=3/5cosb=5/13sin(a-b)=sinacosb-cosasinb=-4/13-36/65=-56/65cos(a-b)/2=（根号下1+sina（a-b））/2

∵tan(π+a)=tana=-1/3,且a是钝角∴sina=√10/10,cosa=-3√10/10又∵a-b是锐角∴cos(a-b)=4/5∴sinb=sin[a-(a-b)]=sinacos(a

cos2a=-5/1390

cosα=-√1-(sinα)^2=-3/5,cosβ=√1-(sinβ)^2=5/13,cos(α-β)=cosαcosβ+sinαsinβ=-3/13+48/65=33/65cos[(α-β)/2

角a为钝角,且sin(a+π/12)=1/3,则a+π/12为钝角,于是cosa(a+π/12)=-（2根2）/3a+5π/12=a+π/12+π/3cos（a+5π/12）=cos（a+π/12）c

A为钝角sinA=2√2/3∴cosA=-1/3余弦定理得cosA=(b²+c²-a²)/(2bc)-2c=9+c²-16c²+2c-7=0c=2√2

不用公式的话过B作BD垂直于CA的延长线于D,连接DAsin∠BAD=sin∠BAC=4/5,∴BD=AB*sin∠BAD=4AB=5,BD=4,所以DA=3BD=4,DC=DA+AC=6,所以BC=

90