两道三角函数题
1.若tanA=2,则3sin²A-sinAcosA-8cos²A=?2.若sinα-cosα=7/13(0°<α<90°),则tanα=?谢谢喽...
1. 若tanA=2,则3sin²A-sinAcosA-8cos²A=?
2. 若sinα-cosα=7/13(0°<α<90°),则tanα=?
谢谢喽 展开
2. 若sinα-cosα=7/13(0°<α<90°),则tanα=?
谢谢喽 展开
2个回答
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1. tanA=2, (secA)^2 = 5, (cosA)^2 = 1/5
3sin²A-sinAcosA-8cos²A = cos²A ( 3 tan²A - tanA -8) = (1/5) ( 12 - 2 -8 ) = 2/5
2. sinα-cosα=7/13(0°<α<90°), (sinα-cosα)^2 = 49/169
(tanα-1) ² = (49/169) (1+ tan²α )
=> ( 5 tanα -12)(12 tanα - 5) =0
=> tanα = 12/5 (sinα>cosα, tanα>1)
3sin²A-sinAcosA-8cos²A = cos²A ( 3 tan²A - tanA -8) = (1/5) ( 12 - 2 -8 ) = 2/5
2. sinα-cosα=7/13(0°<α<90°), (sinα-cosα)^2 = 49/169
(tanα-1) ² = (49/169) (1+ tan²α )
=> ( 5 tanα -12)(12 tanα - 5) =0
=> tanα = 12/5 (sinα>cosα, tanα>1)
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1.解: tanA=2,sin²A+cos²A=1,
3sin²A-sinAcosA-8cos²A=(3sin²A-sinAcosA-8cos²A)/(sin²A+cos²A)分子分母同除以cos²A得
( 3 tan²A -tanA -8)/(tan²A+1)=( 12 - 2 -8 )/(4+1) = 2/5
2.解:由sinα-cosα=7/13(0°<α<90°),两边平方得1 -2sinαcosα=49/169,
∴2sinαcosα=120/169(0°<α<90°),
又(sinα+cosα)^2=1+2sinαcosα=1+120/169=289/169,∴sinα+cosα=17/13
由sinα-cosα=7/13和sinα+cosα=17/13解得sinα=12/13,cosα=5/13,∴tanα = 12/5
3sin²A-sinAcosA-8cos²A=(3sin²A-sinAcosA-8cos²A)/(sin²A+cos²A)分子分母同除以cos²A得
( 3 tan²A -tanA -8)/(tan²A+1)=( 12 - 2 -8 )/(4+1) = 2/5
2.解:由sinα-cosα=7/13(0°<α<90°),两边平方得1 -2sinαcosα=49/169,
∴2sinαcosα=120/169(0°<α<90°),
又(sinα+cosα)^2=1+2sinαcosα=1+120/169=289/169,∴sinα+cosα=17/13
由sinα-cosα=7/13和sinα+cosα=17/13解得sinα=12/13,cosα=5/13,∴tanα = 12/5
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