Question

数学高手求帮忙给好评

数学高手求帮忙给好评

Answer 1

5.
tanx=sinx/cosx=2
cosx=±1/√(1+tan²x)=±√5/5
sinx=±√(1-cos²x)=±2√5/5
7.
1)2(1-sina)(1+cosa)=2-2sina+2cosa-2sinacosa
(1-sina+cosa)²=1-2sina+2cosa+sin²a+cos²a=2-2sina+2cosa-2sinacosa
∴2(1-sina)(1+cosa)=(1-sina+cosa)²
2)sin²a+sin²b-sin²asin²b+cos²acos²b
=sin²a(1-sin²b)+sin²b+cos²acos²b
=sin²acos²b+sin²b+cos²acos²b
=(sin²a+cos²a)cos²b+sin²b
=cos²b+sin²b
=1
8.
1)原式=(4tana-2)/(5+3tana)=5/7
2)原式=tanacos²a=±3*1/(1+tan²a)=±3/10
3)原式=1+2sinacosa=1±3/5

Answer 2

5.(sinx)^2+(cosx)^2=1, 1=4(cosx)^2+(cosx)^2=5(cosx)^2
(cosx)^2=1/5, cosx=1/√5, sinx=2/√5 tanx=2,
或 cosx=-1/√5, sinx=-2/√5 tanx=2,
7.（1）右边=(1-sinx)^2+2(1-sinx)cosx+(cosx)^2
=(1-sinx)^2+2(1-sinx)cosx+1-(sinx)^2
=(1-sinx)^2+2(1-sinx)cosx+(1+sinx)(1-sinx)
=(1-sinx)[1-sinx+2cosx+1+sinx]
=(1-sinx)(2+2cosx)=2(1-sinx)(1+cosx)=左边
（2）.左边=(sinα)^2[1-(sinβ)^2+(cosα)^2(cosβ)^2+(sinβ)^2
=(sinα)^2*(cosβ)^2+(cosα)^2*(cosβ)^2+(sinβ)^2
=(cosβ)^2[(sinα)^2+(cosα)^2]+(sinβ)^2
=(cosβ)^2+(sinβ)^2=1=右边。
8.tanα=3, sinα=3cosα, 1=(sinα)^2+(cosα)2=10(cosα)^2,
(cosα)^2=1/10
(1). 原式=(4tanα-2) / (3tanα+5)=(12-2)/(9+5)=5/7
(2). sinαcosα=(sinα / cosα)(cosα)^2=tanα(cosα)^2=3/10
(3). (sinα+cosα)^2=(sinα)^2+2sinαcosα+(cosα)^2
=1+2*3/10=1+3/5=8/5.