Question

1.已知γ=2β=4α≠0,且α∈[0,2п],且sin²β=sinαsinγ≠0,

Answer 1

(1)

r=2b=4a

sinb^2=sin2a^2=4sina^2cosa^2=4(1-cosa^2)cosa^2

sinαsinγ

=sinasin4a

=sina(2sin2acos2a)

=sina[4sinacosa(2cosa^2-1)]

=4sina^2cosa(2cosa^2-1)

=4(1-cosa^2)cosa(2cosa^2-1)

sin²β=sinαsinγ

4(1-cosa^2)cosa^2=4(1-cosa^2)cosa(2cosa^2-1)

4(1-cosa^2)cosa(cosa-1)(2cosa+1)=0

cosa=1或-1或0或-1/2

由于sin²β=sinαsinγ≠0

所以cosa=-1/2

由于α∈[0,2п]

a=2/3*Pi,b=2a=4/3*Pi,r=4a=8/3*Pi

或者

a=4/3*Pi,b=2a=8/3*Pi,r=4a=16/3*Pi

(2)

3sin²α+2sin ²β=1

3sina^2+2*(1-cos2b)/2=1

3sina^2=cos2b (1)

3sin2α-2sin2β=0

3/2*sin2a=3sinacosa=sin2b (2)

(1)^2+(2)^2得：

9sina^2cosa^2+9sina^4=1

9sina^2(sina^2+cosa^2)=1

sina^2=1/9

由于α、β均为锐角

则sina=1/3

由(1)得：

cos2b=3sina^2=3*(1/9)=1/3

所以cos2b=sina=1/3

所以

α+2β=90

证毕