已知a属于(π 2,π)且sina 2 cosa 2=根号六 2

sin二分之a=五分之四,且a属于(二分之π,π),所以cosa0sina=√1-cos²a=√576/625=24/25

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

f(x)=[sin(x+a/2)+根号3cos(x+a/2)]*cos(x+a/2)=sin(x+a/2)cos(x+a/2)+根号3cos(x+a/2)cos(x+a/2)=1/2sin(2x+a)

（1）∵向量a⊥向量b,∴向量a*向量b=0,即6(sinα)^2+5sinαcosα-4(cosα)^2=0,因式分解得(2sinα-cosα)(3sinα+4cosα)=0,∴tanα=1/2或-

2(sina)^2-sinacosa-3(cosa)^2=0,(2sinα-3)(sinα+1)=0,sinα=-1(sinα=3/2舍去)sin(a+π/4)/(sin2a+cos2a+1)=sin

1.∵a⊥b∴a*b=0∴3sinα*2sinα+5sinαcosα-4cos²α=0∴6tan²α+5tanα-4=0(2tanα-1)(3tanα+4)=0tanα=1/2或-

那么sin^2a/cos^2a的值等于9/16已知sina=3/5,且a属于(π/2,π)所以cosa=-4/5,(sin^2a+cos^2a=1)tana=sina/cosa=-3/4sin^2a/

【参考答案】∵sinA=5/13再问：我不懂为什么sinA/2=√[(1-cosA)/2]？再答：根据半角公式sin(A/2)=±√[(1-cosA)/2]，由于A/2∈(5π/12，π/2)，故取正

五分之二倍根号五五分之根号五四十五度

/>因为sina=3/5,a属于(π/2,π)所以cosa=-4/5(1)cos(a-π/4)=cosa*cosπ/4+sina*sinπ/4=-根号2/10；(2)sin(a/2)的平方+tan(a

方便起见,用a,b来表示α,β由题意得：sina=√2sinb√3cosa=√2cosb两式平方相加得：sin²a+3cos²a=2即：1+2cos²a=2得：cos&#

解.2sin²a-sinacosa-3cos²a=(2sina-3cosa)(sina+cosa)=0∵a∈(0,π/2）∴2sina=3cosa即sina=3/√13,cosa=

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,

1/sinβ=(cosαcosβ-sinαsinβ)sinα整理得:(1+cosα*cosα)sinβ=2sinαcosαcosβ所以,tanβ=sinαcosα/(1+1-2(sinα)^2)/2=

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

π/2

(1)sin(2A+π/6)+sin(2A-π/6)+2cos^2A>=2(sin2Acosπ/6+cos2Asinπ/6)+(sin2Acosπ/6-cos2Asinπ/6)+cos2A+1≥2(√

sin(5π+a)=sin(4π+π+a)=-sina=-3/5sina=3/5a属于（2分之π,π）tana=-sina/√（1-sin^2a)=-(3/5)/(4/5)=-3/4tanβ=1/2t