catty541004
幼苗
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由万能公式得到
cos2A=[1-(tanA)^2]/[1+(tanA)^2]
=(1-4)/(1+4)
=-3/5
tan2A=(2tanA)/[1-(tanA)^2]=4/(1-4)=-4/3
(1)
[(sinA)^2]-sinAcosA+2
=(1/2){2[(sinA)^2]-2sinAcosA+4}
=(1/2)(1-cos2A-sin2A+4)
=(5/2)-(1/2)(cos2A+sin2A)
=(5/2)-(1/2)[-3/5+ (-3/5)(-4/3)]
=(5/2)-(1/2)(1/5)
=12/5
(2)
{[(sinA)^3]-cosA}/(5sinA+3cosA)
={[(sinA)^2]-1/tanA}/[5+(3/tanA)]
={1-[(cosA)^2]-1/2}/[5+(3/2)]
={(1/2)-[(cosA)^2]}/(13/2)
=(2/13){(1/2)-1/[[(tanA)^2]+1]}
=(2/13)[(1/2)-(1/5)]
=(2/13)(3/10)
=3/65
1年前
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