若sinA+cosA/sinA-cosA=2,则sin(A-5π)*sin(3π/2-A)等于?要详细的步骤
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若sinA+cosA/sinA-cosA=2,则sin(A-5π)*sin(3π/2-A)等于?要详细的步骤
若sinA+cosA/sinA-cosA=2,则sin(A-5π)*sin(3π/2-A)等于?
要详细的步骤
若sinA+cosA/sinA-cosA=2,则sin(A-5π)*sin(3π/2-A)等于?要详细的步骤
(sina+cosa)/(sina-cosa)=2
所以sina=3cosa
因为sina*sina+cosa*cosa=1,
所以sina=3*根号10/10,cosa=根号10/10,或者sina=-3*根号10/10,cosa=-根号10/10;
sin(A-5π)*sin(3π/2-A)=(-sinA)*(-sin(A+π/2)=sina*cosa=3/10
http://zhidao.baidu.com/question/112046616.html?si=1
(sina+cosa)/(sina-cosa)=2
sina=3cosa
sina=3/√10 cosa=1/√10
或者 sina=-3/√10 cosa=-1/√10
sin(a-5π)*sin(3π/2-a)
sin(a-5π) =-sina
sin(3π/2-a)=-cosa
sin(a-5π)*sin(3π/2-a)= sina * cosa=3/10
由(sinA+cosA)/(sinA-cosA)=2,知cosA≠0,于是进一步得
(tanA+1)/(tanA-1)=2 解得tanA=3
于是可得
sin(A-5π)*sin(3π/2-A)
=(-sinA)*(-cosA)
=1/2*sin2A
=1/2*2tanA/(1+tan²A)
=tanA/(1+tan²A)
代入tanA=3,可得原式=3/10