高二 数学 数学 请详细解答,谢谢! (9 12:5:52)
1个回答

sin(3π-α)=√2cos(1.5π+β)

sin[2π+(π-α)]=√2cos[π+(π/2)+β]

sin(π-α)=-√2cos[(π/2)+β]

-(-sinα)=-√2(-sinβ)

sinα=√2sinβ

√3cos(-α)=-√2cos(π+β)

√3cosα=√2cosβ

(sinα)²+(cosα)²=2(sinβ)²+(√6/3)²(cosβ)²

=2(sinβ)²+2/3[1-(sinβ)²

=4/3(sinβ)²+2/3

=1

(sinβ)²=1/4

0<β<π

sinβ=1/2

β=π/6或5π/6

sinα=√2sinβ=√2/2

0<α<π

α=π/4或3π/4

故α=π/4,β=π/6或α=3π/4,β=5π/6