已知tanA=2且A属于(0,π 2)则cos2A

(sinA+cosA)^2=1/4sin^2A+cos^2A+2sinAcosA=1/42sinAcosA=-3/4sinAcosA=-3/8sinAcosA/(sin^2A+cos^2A)=-3/8

答：π/4.因为3sinb=3sin[(a+b)-a]=3sin(a+b)cosa-3cos(a+b)sinasin(2a+b)=sin[(a+b)+a]=sin(a+b)cosa+cos(a+b)s

能把题目拍下来吗再问：再问：12题再答：再问：最后一步为什么？再问：a+b不是等于60°吗再答：取值范围不允许再答：a＞90，b＞90再问：再问：ab都在一四象限再答：派等于-派再答：你想想如果两个都

sina+cosa=-7/13①(sina+cosa)²=sin²a+cos²a+2sinacosa=49/169sin²a+cos²a=12sina

tan(A+B)=4(tanA+tanB)/(1-tanAtanB)=4tanAtanB=1/2再联立tanA+tanB=2又tanA

tana+tanb=-3√3tana*tanb=4tan(a+b)=(tana+tanb)/(1-tana*tanb)=√3接下来判断a,b范围,根据tana*tanb=4>0,那么tana和tanb

x^2-5x+6=0x=2,3不妨设tana=2tanb=3所以sina=(2根号5）/5,cosa=根号5/5.sinb=(3根号10）/10,cos=根号10/10cos(a-b)=cosa*co

∵cosa=-1/3,且a属于（π/2,π）∴sina>0,sina=√(1-cos²a)=2√2/3∴tana=sina/cosa=2√2sin2a=2sinacosa=2*2√2/3*(

4tana/2=1-tan^2a/22tana/2/(1-tan^2a/2)=1/2=tanasina=(1/2)/√(1+1/4)=√5/5cosa=√(20/25)=2√5/5sin2a=2*5*

tana=-2所以sina=2/根号5cosa=-1/根号5所以,原式=根号3/2*sina+1/2*cosa=（2根号3-1）/（2根号5）=（2根号15-根号5）/10稍稍说一下：先不管a的象限,

a属于（π,3π/2）第三象限所以sina

tan(a-b)=(tana-tanb)/(1+tanatanb)=1/2又tanb=-1/7∴(tana+1/7)/(1-tana/7)=1/22tana+2/7=1-tana/7tana=(1-2

sina+cosa=tana-1/tanasina+cosa=sina/cosa-cosa/sinasina+cosa=(sin²a-cos²a)/sinacosa1=(sina-

∵tana、tanb是方程x²-5x+6=0的两个根且方程的解为x1=2,x2=3∴tana=2,tanb=3（或tana=3,tanb=2）∵a,b属于（0,π）且tana、tanb都大于

已知a,b属于（0,π）,且tana、tanb是方程x²-5x+6=0的两个根,求tan(a+b)的值.易得tana+tanb=5,tana*tanb=6tan(a+b)=(tana+tan

∵sin²(2a)+sin(2a)cosa-cos(2a)=1∴sin²(2a)+sin(2a)cosa-cos(2a)-1=0==>sin²(2a)+sin(2a)co

tana=3tanb,且a,b属于[0,π/2）,a>b,tanb>0tan(a-b)=(tana-tanb)/(1+tanatanb)=2tanb/(1+3tan²b)1+3tan&sup

由已知条件,得(tanA-tanB)/(tanAtanB)=2tanA1/tanB-1/tanA=2tanA即cotB-cotA=2tanAcotB=2tanA+cotAsin(2A+B)/sinB=

解；：(tana/2)/(1-tana/2的平方)=1/4∴2tana/2/(1-tan²a/2)=1/2∴tana=1/2∵a,b∈(0,π/4)∴2sina=cosa∴sina=√5/5

是(sina+cosa)/(sina-cosa)还是sina+cosa/sina-cosa无括号?是(sina+cosa)/(sina-cosa)的话=[(sina+cosa)/cosa]/[(sin