已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana...
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![已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana...](/uploads/image/z/2666954-2-4.jpg?t=%E5%B7%B2%E7%9F%A5sin%5E2+2a%2Bsin2acosa-cos2a%3D1%2Ca+E%280%2C%CF%80+%2F2%29+%E6%B1%82sina+tana...)
已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana...
已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana
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已知sin^2 2a+sin2acosa-cos2a=1,a E(0,π /2) 求sina tana...
sin^2 2a=(2sinacosa)^2=4(sina)^2*(cosa)^2
sin2acosa=2sina(cosa)^2
cos2a=2(cosa)^2-1
sin^2 2a+sin2acosa-cos2a
=4(sina)^2*(cosa)^2+2sina(cosa)^2-[2(cosa)^2-1]
=1
所以4(sina)^2*(cosa)^2+2sina(cosa)^2-2(cosa)^2=0
而a E(0,π /2),cosa≠0,
所以4(sina)^2+2sina-2=0,
解得sina=1/2或者-1(舍去)
所以cosa=(根号3)/2
tana==(根号3)/3
∵sin²(2α)+sin(2α)cosα-cos(2α)=1
∴sin²(2α)+sin(2α)cosα-cos(2α)-1=0
==>sin²(2α)+sin(2α)cosα-2cos²α=0
==>[sin(2α)+2cosα][sin(2α)-cosα]=0...
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∵sin²(2α)+sin(2α)cosα-cos(2α)=1
∴sin²(2α)+sin(2α)cosα-cos(2α)-1=0
==>sin²(2α)+sin(2α)cosα-2cos²α=0
==>[sin(2α)+2cosα][sin(2α)-cosα]=0
==>(2sinαcosα+2cosα)(2sinαcosα-cosα)=0
==>2cos²α(sinα+1)(2sinα-1)=0
∵α∈(0,90°)
∴cosα≠0,sinα+1≠0
∴2sinα-1=0 ==>sinα=1/2
==>α=30°
==>tanα=√3/3
故sinα=1/2,tanα=√3/3
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