Question

已知向量a=（cosa,sina）,向量b(cos²a,sin²a),且向量a⊥向量b，则向量b=？

Answer 1

1.
取内积得
cos(a)cos^2(a)+sin(a)sin^2(a)=0，所以
[cos(a)+sin(a)][cos^2(a)-cos(a)sin(a)+sin^2(a)]=0
不过cos^2(a)-cos(a)sin(a)+sin^2(a)=1-(1/2)sin(2a)>0
所以
cos(a)+sin(a)=0
即
cos(a)=-sin(a)
所以
cos(a)=-sin(a)=±1/√2
所以向量b=(1/2,
1/2)
2.
p+q=sin(a)sin(b)+cos^2[(a+b)/2]
=(1/2)[cos(a-b)-cos(a+b)]+(1/2)[1+cos(a+b)]
【积化和差、半角公式】
=(1/2)[cos(a-b)+1]
所以
0≤p+q≤1
3.
sin(7π/3)*cos(-11π/6)+tan(-15π/4)*1/tan(13π/6)
=sin(π/3)*cos(π/6)+tan(π/4)*1/tan(π/6)
=(√3)/2
*
(√3)/2
+
1*1/(1/√3)
=2√3
4.
每个三棱锥的体积是
(1/6)*(1/2)*(1/2)*(1/2)=1/48
所以截去了体积1/6
所以剩下了体积5/6的截半立方体

Answer 2

1.得到cos^3α+sin^3α=0
分解因式：(cosα+sinα)(cos^2α-cosαsinα+sin^2α)=0
得：cosα+sinα=0或cos^2α-cosαsinα+sin^2α=0
1)
-cosα=sinα
tanα=-1
α=225°
得b(cos^2
225°,sin^2
225°)=(1/2,1/2)
2)cos^2
α-cosαsinα+sin^2
α=0
1-cosαsinα=0
sin2α/2=1
sin2α=2
无解
综上，b=(1/2,1/2)
2.p+q
=cos^2[(α+β)/2]+sinαsinβ
=[1+cos(α+β)+2sinαsinβ]/2
=(1+cosαcosβ-sinαsinβ+2sinαsinβ)/2
=(1+cosαcosβ+sinαsinβ)/2
=[1+cos(α-β)]/2
cos(α-β)∈[-1,1]
1+cos(α-β)∈[0,2]
[1+cos(α-β)]/2∈[0,1]
即p+q∈[0,1]
3.sin(7π/3)*cos(-11π/6)+tan(-15π/4)*1/tan(13π/6)
=sin(2π+π/3)*cos(-2π+π/6)+tan(-4π+π/4)/tan(2π+π/6)
=sin(π/3)*cos(π/6)+tan(π/4)/tan(π/6)
=(√3)/2*(√3)/2+1/[(√3)/3]
=2√3
4.设棱长为2a，总体积为8a^3
每个三棱锥体积为1/3*(1/2*a*a)*a=(a^3)/6
一共8个，体积为4(a^3)/3
所以余下的体积为(8-4/3)a^3=(20/3)a^3
(20/3)a^3/(8a^3)=5/6