Show that sin3x/sinx = 2cos2x + 1
thanks ~
amaths ~
2008-12-07 4:08 am
回答 (2)
2008-12-07 4:17 am
✔ 最佳答案
LHS=sin3x/sinx
=(3sinx-4sin^3x)/sinx
=3-4sin^2x
=3-4(1-cos^2x)
=3-4+4cos^2x
=4cos^2x-1
=4[(1+cos2x)/2]-1
=2cos2x + 1
=RHS
2008-12-07 5:18 am
LHS
sin3x/sinx
= [sin(2x x)]/sinx
= [sin2xcosx+sinxcos2x]/sinx
= [2sinxcosxcosx+(2cosxcosx-1)sinx]/sinx
= 2cosxcosx+2cosxcosx-1
= 2cosxcosx+cos2x
= 2cosxcosx+cosx-1+1
= cos2x+cos2x+1
= 2cos2x+1
= RHS
(I have to paste my answer here 'cuz I accidentally deleted my answer)
sin3x/sinx
= [sin(2x x)]/sinx
= [sin2xcosx+sinxcos2x]/sinx
= [2sinxcosxcosx+(2cosxcosx-1)sinx]/sinx
= 2cosxcosx+2cosxcosx-1
= 2cosxcosx+cos2x
= 2cosxcosx+cosx-1+1
= cos2x+cos2x+1
= 2cos2x+1
= RHS
(I have to paste my answer here 'cuz I accidentally deleted my answer)
收錄日期: 2021-04-25 16:58:19
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