＜急＞求救數學 [復制鏈接]

*=度
請做埋步驟比我ｐｌｅａｓｅ
１．苦sinθ= - 5/13 及270*<θ<306*,求sinθ-2cosθ/2sinθ-cosθ的值

係唔係270-360呀?

draw a CAST diagram, find all the 3 sides, x,y and r, then should be solved easily.

下面是引用kylau於2006-04-23 12:02發表的:
draw a CAST diagram, find all the 3 sides, x,y and r, then should be solved easily.

你係指求左果３條邊先呀？
ｅ個我知
都係唔知點計

sinθ=y/r
cosθ=x/r
x^2+y^2=r^2

再用CAST睇正負
揾到cosθ咪得lor......

cosθ = 開方[1- (5/13)^2] = 12/13
係未咁做？

下面是引用cool於2006-04-23 12:53發表的:
cosθ = 開方[1- (5/13)^2] = 12/13
係未咁做？

個方法有d怪.......


但係又o岩.....

下面是引用傷心小sam於2006-04-23 12:57發表的:
個方法有d怪.......
[表情]
但係又o岩..... [表情]

急急急急知
請做ｅ條數既步驟出黎ｐｌｚ

下面是引用cool於2006-04-23 13:00發表的:
急急急急知 [表情]
請做ｅ條數既步驟出黎ｐｌｚ [表情]

1.畫個直角三角形先
2.let r同x之間個angle做@(<90*)
3.sin@=y/r=5/13 ar ma
  y=5,r=13 lor...
  用pyth thm揾埋x
4.cos@=x/r
5.用CAST分正負你識ga la~

1. (tan A - tan B)/(1 + tan A tan B)
= [(-5/12) - (-3/4)]/[1 + (-5/12)(-3/4)]
= (-5+9/12) /1+(-5/16)
=(4/12) (16/11)
=16/33
我想問邊度計錯左呀？？？？？

下面是引用cool於2006-04-23 17:45發表的:
1. (tan A - tan B)/(1 + tan A tan B)
= [(-5/12) - (-3/4)]/[1 + (-5/12)(-3/4)]
= (-5+9/12) /1+(-5/16)
=(4/12) (16/11)
=16/33
.......

(-5/12)(-3/4)=5/16..=,=

下面是引用cool於2006-04-23 17:45發表的:
1. (tan A - tan B)/(1 + tan A tan B)
= [(-5/12) - (-3/4)]/[1 + (-5/12)(-3/4)]
= (-5+9/12) /1+(-5/16)
=(4/12) (16/11)
=16/33
.......

負負得正

自己check清楚先la........

cos(-θ））{1+sin^(180-θ)/1+cos^(180+θ)+sin^(-θ)

下面是引用cool於2006-04-23 17:51發表的:
cos(-θ））{1+sin^(180-θ)/1+cos^(180+θ)+sin^(-θ)

做乜ar.......

下面是引用傷心小sam於2006-04-23 17:49發表的:
負負得正
自己check清楚先la........ [表情]

ｅ∼∼係喎