已知cos(π/2-α)-sin(π/2+α)=1/5,且0<α<π,求下列各式的值:
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已知cos(π/2-α)-sin(π/2+α)=1/5,且0<α<π,求下列各式的值:
(1)sin(π+α)cos(π+α);;(2)sin(3π/2-α)+cos(3π/2-α);;
(3)cos3^(3π/2+α)-sin^3(3π/2+α)
(1)sin(π+α)cos(π+α);;(2)sin(3π/2-α)+cos(3π/2-α);;
(3)cos3^(3π/2+α)-sin^3(3π/2+α)
/>cos(π/2-α)-sin(π/2+α)=sina-cosa=1/5 ①
两边平方得:
1-2sinacosa=1/25
sinacosa=12/25
∴sin^2 a+cos^2 a=(sina+cosa)^2-2sinacosa
=(sina+cosa)^2-24/25
=1
∴(sina+cosa)^2=49/25
又 0<α<π
∴ sina+cosa=7/5②
由①②得sina=4/5 cosa=3/5
(1) : sin(π+α)cos(π+α)
=sinacosa
=12/25
(2): sin(3π/2-α)+cos(3π/2-α)
=-cosa-sina
=-(sina+cosa)
=-7/5
(3) cos3^(3π/2+α)-sin^3(3π/2+α)
=sin^3 a+cos^3 a
=(sina+cosa)(1-sinacosa)
=(7/5)*(1-12/25)
=91/125
两边平方得:
1-2sinacosa=1/25
sinacosa=12/25
∴sin^2 a+cos^2 a=(sina+cosa)^2-2sinacosa
=(sina+cosa)^2-24/25
=1
∴(sina+cosa)^2=49/25
又 0<α<π
∴ sina+cosa=7/5②
由①②得sina=4/5 cosa=3/5
(1) : sin(π+α)cos(π+α)
=sinacosa
=12/25
(2): sin(3π/2-α)+cos(3π/2-α)
=-cosa-sina
=-(sina+cosa)
=-7/5
(3) cos3^(3π/2+α)-sin^3(3π/2+α)
=sin^3 a+cos^3 a
=(sina+cosa)(1-sinacosa)
=(7/5)*(1-12/25)
=91/125
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