cos2a/sin(a-π/4) =-根号（2）/2,则cosa+sina的值为?

1. cos2a/sin(a-π/4)=-√2/2,那么sina+cosa的值为?
cos2a=(-√2/2)(sinacosπ/4-cosasinπ/4)=(-sina-cosa)/2
cos2a=cos^2a-sin^2a=(-sina-cosa)/2
cos^2a-cosa/2=sin^2a-sina/2
(cosa-1/4)^2=(sina-1/4)^2
所以cosa=sina=√2/2,cosa+sina=√2或cosa-1/4=1/4-sina,cosa+sina=1/2
2.在三角形ABC中,tanA=1/2,cosB=3√10/10,则tanC的值是?
tanA=1/2,2sinA=cosA,4sin^2A=cos^2A,sinA=√5/5,cosA=2√5/5
cosB=3√10/10,sinB=√10/10
sinC=sin(B+A)=sinBcosA+sinAcosB=√2/2
cosC=-cos(B+A)=-(cosBcosA-sinBsinA)=-√2/2
tanC=-1
3. 要使得sina-√3cosa9=(4m-6)/(4-m)有意义,则m的取值范围是?
sina-√3cosa9?sina-√3cosa?
-2

cos2a/sin(a-π/4)=-√2/5
cos2a*cos(a-π/4)/sin(a-π/4)*cos(a-π/4)
=2cos2a*cos(a-π/4)/sin(2a-π/2)
=-2cos2a*cos(a-π/4)/cos2a
=-2cos(a-π/4)=-√2/5
cos(a-π/4)=√2/10
(1)cosa+sina
=√2cos(a-π/4)
=1/5
(2)sina-cosa
=√2sin(a-π/4）
因为cos(a-π/4)=√2/10
- pai

∵sina=1/2+cosa
∴sina-cosa=1/2
∴(sina-cosa)²=1/4
∴1-2sinacosa=1/4
∴2sinacosa=3/4
∴(cosa+sina)²
=1+2sinacosa
=7/4
∵a∈（0,π/2）
∴cosa+sina>0
∴cosa+sina=√7/2
∴cos2a/sin（a-π/4）
=(cos²a-sin²a)/[√2/2cosa-2/2sina]
=√(cosa+sina)(cosa-sina)/[√2/2(cosa-sina)]
=√2(sina+cosa)
=√14/2
何意,提交了,你补充说不要粘贴的,放心,
我就是靠数学吃饭的

sina=1/2+cosa=>sina-cosa=1/2
a∈（0,2/π）=>cosa>0=>sina>1/2
上式平方=>1-sin2a=1/4=>sin2a=3/4a∈（0,2/π）=>2a∈（0,4/π）=>cos2a>0
=>cos2a=根号7/4
sin（a-π/4）=根号2(sina-cosa)/2=根号2/4
cos2a/sin（a-π/4）=根号7/4/（根号2/4）=根号(7/2)

sin(π/4＋a)
=sin(π/2＋a-π/4)
=cos(a-π/4)
=cos(π/4-a)
=12/13.
因为cos(π/4-a)=12/13,
所以由二倍角公式得
cos(π/2-2a)=sin2a
=2cos(π/4-a)^2-1
=2*(12/13)^2-1
=119/169.
因为0

sin(a)-cos(a)=√2sin(a-π/4)=√2cos(a+π/4)=-2sin(2a)
sin(a)+cos(a)=√2cos(a-π/4)=√2sin(a+π/4)
4sin(a)cos(a)=1-cos(2a+π/2)-2sin^2(2a)=1+sin(2a)-4sin^2(2a)=2sin(2a)
4sin^2(2a)+sin(2a)-1=0
sin(2a)=(-1+_√17)/8
sin(a-π/4)=(1_+√17)/4
sin(a+π/4)=-cos(2a)=懒得算
最后看看象限的范围吧

∵sina=1/2+cosa
∴sina-cosa=1/2
∴(sina-cosa)²=1/4
∴1-2sinacosa=1/4
∴2sinacosa=3/4
∴(cosa+sina)²
=1+2sinacosa
=7/4
∵a∈（0,π/2）
∴cosa+sina>0
∴cosa+sina=√7/2
∴cos2a/sin（a-π/4）
=(cos²a-sin²a)/[√2/2cosa-2/2sina]
=√(cosa+sina)(cosa-sina)/[√2/2(cosa-sina)]
=√2(sina+cosa)
=√14/2
何意,提交了,你补充说不要粘贴的,放心,
我就是靠数学吃饭的

sin(a+pai/4)=√2/2(sina+cosa)
cos2a=cos^2a-sin^2a=(cosa+sina)*(cosa-sina)
cos2a/sin(a+pai/4)=√2/2=>cosa-sina=1/2
两边平方得：sin2a=-3/4

由cos2α/sin(α-π/4)=-(√2)/2,
且
cos2α=cos²α-sin²α
sin(α-π/4)=(√2)/2*(sinα-cosα)
则:
cos2α/sin(α-π/4)=cos²α-sin²α/((√2)/2*(sinα-cosα))=-(√2)/2
等式左边同时约去sinα-cosα,可得:
√2(sinα+cosα)=-(√2)/2
则有:sinα+cosα=-1/2

（sina+cosa）^2+（sina-cosa）^2=2,展开不就是了吗,孩子,数学要好好学.

cos2a/sin（a-π/4）=(cos^a-sin^a)/[√2/2(sina-cosa)]=-√2(sina+cosa)
设(sina+cosa)=t,题设sina-cosa=1/2,则sina=(t+1/2)/2,cosa=(t-1/2)/2,
因为(sina)^2+(cosa)^2=1,求出t=√7/2
y=-√14/2

cos2a/sin(a-pai/4)=-根号2／2
cos2a/(sinacos(pai/4)-cosasin(pai/4))=-根号2／2
cos2a/(根号2／2*sina-根号2／2*cosa)=-根号2／2
cos2a/根号2／2*(sina-cosa)=-根号2／2
(cos^2 a-sin^2 a)/(sina-cosa)=-1/2
(cosa-sina)(cosa+sina)/(sina-cosa)=-1/2
cosa+sina=1/2.

cos2a/sin（a-π/4)= -√2/2,
sin(π/2-2a) /sin（a-π/4)= -√2/2,
-sin(2a-π/2) /sin（a-π/4)= -√2/2,
[-2 sin（a-π/4)cos（a-π/4)） /sin（a-π/4)= -√2/2,
-2 cos（a-π/4)）= -√2/2,
-2[cos a cosπ/4+sin a sinπ/4] = -√2/2,
-√2(cosa+sina) = -√2/2,
所以cosa+sina=1/2.

cos2a=cos^a-sin^a=(cosa+sina)(cosa-sina)=√2/2*sin(a-π/4)=√2/2（sina*√2/2-cosa√2/2)=1/2(sina-cosa)
所以cosa+sina=-1/2

sin(2a)/sin(π+a) +[cos(2a)+1]/cosa
=2sinacosa/(-sina)+ 2cos²a/cosa
=-2cosa+2cosa
=0

sina=1/2+cosa =>sina-cosa=1/2
a∈（0,2/π）=>cosa>0 =>sina>1/2
上式平方=>1-sin2a=1/4 =>sin2a=3/4 a∈（0,2/π）=>2a∈（0,4/π）=>cos2a>0
=>cos2a=根号7/4
sin（a-π/4）=根号2(sina-cosa)/2=根号2/4
cos2a/sin（a-π/4）=根号7/4/（根号2/4）=根号(7/2)

因为cos2a=(cosa)^2-(sina)^2
sin（a-45度)=(√2/2sina-√2/2cosa)
即cos2a/sin（a-45度)=[(cosa)^2-(sina)^2]/(√2/2sina-√2/2cosa)=-√2(sina+cosa)=√2/2
所以cosa+sina=-1/2

cos2a/sin(a-派/4)=-根号2/2,
(cosa-sina)(cosa+sina)/[(√2/2)(sina-cosa)]=-√2/2
cosa+sina=-1/2
sin2a=(sina+cosa)²-1=1/4-1=-3/4

通过sina=1/2+cosa和cos²a+sin²a=1解出sina和cosa即可

1、
cos[(π/4)-α]=12/13
===> cos[(π/2)-2α]=2cos²[(π/4)-α]-1
===> cos[(π/2)-2α]=2*(12/13)²-1=119/169
===> sin2α=119/169
===> cos2α=√(1-sin²2α)=120/169
因为[(π/4)-α]+[(π/4)+α]=π/2
所以,sin[(π/4)+α]=cos[(π/4)-α]=12/13
则,原式=(120/169)/(12/13)=10/13
2、
sin[α+(π/3)]=-4/5
则,cos[α+(π/3)]=-3/5
所以,cos[2α+(2π/3)]=2cos²[α+(π/3)]-1=2*(-3/5)²-1=-7/25
那么,sin[2α+(2π/3)]=-√[1-cos²(2α+2π/3)]=-24/25
则,cos[2α+(5π/6)]=cos[2α+(2π/3)+(π/6)]
=cos[2α+(2π/3)]cos(π/6)-sin[2α+(2π/3)]sin(π/6)
=(-7/25)*(√3/2)-(-24/25)*(1/2)
=(24-7√3)/50