maths

2007-05-18 3:22 am
1.Prove that 2sinxcosx indentity to (sinx+cosx)2次方 - 1
2.a)sin(90-x) sinx/(sinx-1)+cos(90-x)cosx/(sinx+1) indentity to -2sinx/tan(90-x)
b)((cosx tanx+cos( 90 - x ) tan( 90 - x ) )2次方-2sinxsin(90-x) indentity to 1


THX!

回答 (3)

2007-05-18 4:10 am
✔ 最佳答案
RHS
=(sinx+cosx)^2-1
=sin^2 x+2sinx cosx+cos^2x-1
=sin^2 x+cos^2 x+2sinx cosx-1
=1+2sinx cosx-1
=2sinx cosx
=LHS//

LHS
=sin(90-x) sinx/(sinx-1)+cos(90-x)cosx/(sinx+1)
=cosx sinx/(sinx -1)+sinxcosx/(sinx+1)
=[(sinx+1)(sinx cosx)+(sinx-1)(cosx sinx)]/[(sinx+1)(sinx-1)]
=(sin^2 x cosx+sinx cosx+sin^2 x cosx-sinx cosx)/(sin^2-1)
=(2sin^2 x cosx)/(-cos^2 x)
=-2tan^2 x cosx
=-2(sin^2 x/cos^2 x)(cosx)
=-2sin^2/cosx
RHS
=-2sinx/tan(90-x)
=-2sinx tanx
=-2sinx(sinx/cosx)
=-2sin^2/cosx
=LHS//

[(cosx tanx+cos(90-x)tan(90-x)]^2-2sinxsin(90-x)
=(sinx+sinx/tanx)^2-2sinx cosx
=[sinx+sinx(cosx/sinx)]^2-2sinx cosx
=(sinx +cosx)^2-2sinx cox
=sin^2 x+2sinx cosx+cos^2 x-2sinx cosx
=sin^2 x+cos^2 x
=1
=RHS
參考: ME
2007-05-18 4:24 am
1.
2sinxcosx
=2sinxcosx + 1 - 1
=2sinxcosx + sin^2x + cos^2x - 1
=sin^2x + 2sinxcosx + cos^2x - 1
=(sinx+cosx)^2 - 1

2(a)
sin(90-x) sinx/(sinx-1)+cos(90-x)cosx/(sinx+1)
=sinxcosx/(sinx-1)+sinxcosx/(sinx+1)
=(sin^2xcosx+sinxcosx+sin^2xcosx-sinxcosx)/(sin^2x-1) ..通分
=2sin^2xcosx/(-cos^2x)
=2sin^2x/(-cosx)
= -2sinx(sinx/cos)
= -2sinxtanx
= -2sinx/tan(90-x)
(b)
[(cosx tanx+cos( 90 - x ) tan( 90 - x ) ]^2 - 2sinxsin(90-x)
=(sinx+sinx/tanx)^2 -2sinxcosx
=(sinx+cosx)^2 - 2sinxcosx
=sin^2x + 2sinxcosx + cos^2x - 2sinxcosx
=sin^2x + cos^2x
=1
參考: 自己計
2007-05-18 4:20 am
1)
(sinx + cosx)^2 - 1
= (sinx)^2 + 2sinxcosx + (cosx)^2 - 1
= 2sinxcosx + 1 - 1
= 2sinxcosx

2a)
sin(90-x) sinx / (sinx - 1) + cos(90-x) cosx / (sinx + 1)
= [ cosx sinx (sinx + 1) ] / [ (sinx + 1) (sinx - 1) ]
+ [ sinx cosx (1 - sinx) ] / [ (1 - sinx) (sinx + 1) ]
= [ cosx sinx (sinx + 1) ] / [ - (cosx)^2 ]
+ [ sinx cosx (1 - sinx) ] / (cosx)^2
= [ - sinx (sinx + 1) + sinx (1 - sinx) ] / cosx
= - 2 (sinx)^2 / cosx
= - 2 sinx tanx

-2 sinx / tan(90-x)
= -2 sinx tanx

L.H.S = R.H.S.

2b)
[ (cosx tanx + cos(90-x) tan(90-x) ]^2 - 2sinxsin(90-x)
= [ sinx + sinx / tanx ]^2 - 2sinxcosx
= [ sinx + cosx ]^2 - 2sinxcosx
= 2sinxcosx + 1 - 2sinxcosx ------------------ by(1)
= 1


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