1、已知A、B是锐角,且sinA=√5／5,sinB=√10／10.求A+B的值

1.∵A、B是锐角
∴cosA=√[1-(sinA)^2]=2√5/5 ,cosB=√[1-(sinB)^2]=3√10/10
cos(A+B)=cosAcosB-sinAsinB=√2/2
A+B=45º
2.sinA+cosA=√2sin(A + π/4)=tanA
∵0＜A＜π/2
∴π/4＜A+π/4＜3π/4
则√2/2＜sin(A+π/4)＜1
∴1＜tanA＜√2
π/4＜tanA＜arctan√2 ,约为（π/4,π/3） ,选C
3.tan(A-B)=(tanA-tanB)/(1+tanA*tanB)
1 - [(tanA-tanB)/(1+tanA*tanB)]/tanA = (sinC)^2/(sinA)^2
[(tanA)^2*tanB+tanB] / [tanA(1+tanA*tanB)] = (sinC)^2/(sinA)^2
tanB*(secA)^2 / tanA(1+tanA*tanB) = (sinC)^2/(sinA)^2 ,(其中1 + tan^2A=sec^2A)
tanB*(secA)^2*(sinA)^2 = tanA(1+tanA*tanB)*(sinC)^2
tanB*tanA = (1+tanA*tanB)*(sinC)^2 ,（两边除以tanA）
(tanB*tanA+1) - 1 = (1+tanA*tanB)*(sinC)^2
1 - 1/(1+tanA*tanB) = (sinC)^2
1 - (sinC)^2 = 1/(1+tanA*tanB)
(cosC)^2 = 1/(1+tanA*tanB)
1/(secC)^2 = 1/(1+tanA*tanB)
(secC)^2 = 1+ tanA*tanB
(secC)^2-1=tanA*tanB
(tanC)^2=tanA*tanB
4.原式=[sinAcos(π/6)+cosAsin(π/6)]^2+[sinAcos(π/6)-cosAsin(π/6)]^2-(sinA)^2
=[(√3/2)sinA + (1/2)cosA]^2 + [(√3/2)sinA - (1/2)cosA]^2 - (sinA)^2
=3(sinA)^2/2 + (cosA)^2/2 - (sinA)^2=1/2*[(sinA)^2 + (cosA)^2]
=1/2
5.①原式=2sin20°cos20°cos40°cos80°/ 4sin20°
=sin40°cos40°cos80°/ 4sin20°
=sin80°cos80°/ 8sin20°
=sin160°/ 16sin20°
=sin(180°-20°)/16sin20°
=1/16
②原式=sin(90°-24°)sin(90°-48°)sin6°sin(90°-12°)
=cos24°cos48°sin6°cos12°
=2sin6°cos6°cos12°cos24°cos48°/2cos6°
=sin12°cos12°cos24°cos48°/2cos6°
=sin24°cos24°cos48°/4cos6°
=sin48°cos48°/8cos6°
=sin96°/16cos6°
=1/16
③原式=sin67.5°/cos67.5° - sin22.5°/cos22.5°
=cos22.5°/sin22.5° - sin22.5°/cos22.5°
=[(cos22.5°)^2 - (sin22.5°)^2] /(sin22.5°*cos22.5°)
=cos45°/ [(1/2)*sin45°]
=2
④原式=1/2 * [cos(5π/12 + π/12) + cos(5π/12 - π/12)]
=1/2 * [cos(π/2)+cos(π/3)]
=1/2 * [0 + 1/2]
=1/2 * 1/2
=1/4
6.原式=[(1+sinA-cosA)^2+(1+sinA+cosA)^2] / (1+sinA+cosA)(1+sinA-cosA)
展开,整理后=4(1+sinA) / 2sinA(sinA+1)
=2/sinA
7.sin(π/4+A)sin(π/4-A)=(-1/2){cos[(π/4+A)+(π/4-A)] - cos[(π/4+A)-(π/4-A)]}
=(-1/2)[cos(π/2) - cos2A] =(1/2)cos2A=1/6
cos2A=1/3
∵π/2＜A＜π
∴π＜2A＜2π
sin2A=-√[1-(cos2A)^2]=-2√2/3
sin4A=2sin2A*cos2A=-4√2/9
8.tan2B=2tanB / 1 - (tanB)^2 =3/4
tan(A+2B)=(tanA + tan2B) / (1 - tanAtan2B) =1
∵A、B都为锐角
∴A+2B∈(0,270°)
A+2B=45°或A+2B=225°
9.原式=(1 - cos40°)/2 + (1 + cos160°)/2 + √3/2*(sin100°-sin60°)
=1 + (cos160°- cos40°)/2 + √3/2*sin100°- 3/4
=1 - sin[(160°+ 40°)/2]*sin[(160°- 40°)/2] + √3/2*sin100°- 3/4
=1 - sin100°sin60° + √3/2*sin100°- 3/4
=1/4
10.原式=[2sin50° + sin10°(cos10°+√3sin10°)/cos10°]×√2sin80°
= [2sin50°cos10° + 2sin10°(cos60°cos10°+sin60°sin10°)×√2sin80°/cos10°
=2[sin50°cos10° + sin10°cos(60°-10°)]×√2
= 2sin(50°+10°)×√2
= √6
11.原式=cos[2(π/3+A)]
=2[cos(π/3+A)]^2 - 1
=2{sin[π/2-(π/3+A)]}^2 - 1
=2[sin(π/6-A)]^2 - 1
=-7/9
12.tan(π/4+A) = [tan(π/4)+tanA] / [1-tan(π/4)tanA] =3
解得：tanA=1/2
sin2A - 2(cosA)^2=sin2A - (1+cos2A) =sin2A - cos2A - 1
=(2tanA)/[1+(tanA)^2] - [1-(tanA)^2]/[1+(tanA)^2] - 1
=-4/5
13.∵在△ABC中,B=π/3
∴A+C=2π/3 则(A+C)/2 =π/3
∵tan(A/2 + C/2) =[tan(A/2) + tan(C/2)] / [1 - tan(A/2)tan(C/2)] =tan(π/3)=√3
∴tan(A/2) + tan(C/2) = √3*[1 - tan(A/2)tan(C/2)]
√3*[1 - tan(A/2)tan(C/2)] = √3 - √3*tan(A/2)tan(C/2)
即：tan(A/2) + tan(C/2) + √3*tan(A/2)tan(C/2) = √3
14.原式=√[2(cos5°)^2]=(√2)cos5°
15.cos(A-π/6) + sinA = cosAcos(π/6) + sinAsin(π/6) + sinA
=cosAcos(π/6) + (1/2)sinA + sinA
=(√3/2)cosA + (3/2)sinA =4√3／5
则(1/2)cosA + (√3/2)sinA = 4/5
即：sin(A + π/6)=4/5
sin(A + 7π/6)=sin[π+(A + π/6)]=-sin(A + π/6)=-4/5
16.题目是不是应该“已知3sinB=sin(2A+B) ,求证：tan(A+B)=2tanA”这样呀?如果是的话,证明如下：
3sinB=sin(2A+B)
3sin[(A+B)-A] = sin[(A+B)+A]
3[sin(A+B)cosA - cos(A+B)sinA] = sin(A+B)cosA + cos(A+B)sinA
3sin(A+B)cosA - 3cos(A+B)sinA = sin(A+B)cosA + cos(A+B)sinA
2sin(A+B)cosA = 4cos(A+B)sinA
tan(A+B)=2tanA