已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2(11π/6-x)
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已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2(11π/6-x)
如题,
如题,
(5π/6-x)+(x+π/6)=π
(5π/6-x)=π-(x+π/6)
sin(5π/6-x)=sin[π-(x+π/6)]=sin(x+π/6)=1/4
(11π/6-x)+(x+π/6)=2π
(11π/6-x)=2π-(x+π/6)
sin(11π/6-x)=sin[2π-(x+π/6)]= - sin(x+π/6)= - 1/4
sin^2(11π/6-x)=1/16
原式=(1/4)+(1/16)=5/16
(5π/6-x)=π-(x+π/6)
sin(5π/6-x)=sin[π-(x+π/6)]=sin(x+π/6)=1/4
(11π/6-x)+(x+π/6)=2π
(11π/6-x)=2π-(x+π/6)
sin(11π/6-x)=sin[2π-(x+π/6)]= - sin(x+π/6)= - 1/4
sin^2(11π/6-x)=1/16
原式=(1/4)+(1/16)=5/16
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