化簡：
1. -(3sin²x+3cos²x)
2. 2cosx over cos(90º-x) ×tanx
f.3 maths ...
2007-02-02 2:50 am
回答 (6)
2007-02-02 3:05 am
✔ 最佳答案
1. -(3sin²x+3cos²x)= -[3sin²x+3(1-sin²x)]-------{sin²+cos²x=1, cos²x=1-sin²x}
= -[3sin²x+3-3sin²x)]
= -3
2. 2cosx over cos(90º-x) × tanx
= [2cosx/cos(90º-x)] × (sinx/cosx)---------{tanx=sinx/cosx}
= (2cosx/sinx) × (sinx/cosx)-----------------{cosx(90º-x)=sinx}
= 2
2007-02-03 1:16 am
1.
=-3
2.
=2
=-3
2.
=2
2007-02-02 3:51 am
►
1.化簡 -( 3sin²x + 3cos²x )
= -3 ( sin²x + cos²x )
= -3 (1)
= -3
►
2.化簡 2cosx / cos( 90º-x ) ( tan x )
= (2cos x / sin x ) ( sin x / cos x )
= 2
►
公式 :
sin²x + cos²x=1
cos( 90º-x ) = sin x
tan x = sin x / cos x
1.化簡 -( 3sin²x + 3cos²x )
= -3 ( sin²x + cos²x )
= -3 (1)
= -3
►
2.化簡 2cosx / cos( 90º-x ) ( tan x )
= (2cos x / sin x ) ( sin x / cos x )
= 2
►
公式 :
sin²x + cos²x=1
cos( 90º-x ) = sin x
tan x = sin x / cos x
2007-02-02 2:58 am
1. -(3sin²x+3cos²x)
= -3(sin²x+cos²x)
= -3(1)
= -3
2. 2cosx / cos(90º-x) ×tanx
= 2cosx / sinx x tanx
= 2cosx / sinx x (sinx / cosx)
= 2
= -3(sin²x+cos²x)
= -3(1)
= -3
2. 2cosx / cos(90º-x) ×tanx
= 2cosx / sinx x tanx
= 2cosx / sinx x (sinx / cosx)
= 2
參考: 自己
2007-02-02 2:55 am
1. -(3sin²x+3cos²x)
= -3(sin²x+cos²x)
= -3(1)
= -3
2. 2cosx over cos(90º-x) ×tanx
= 2cosx/sinx ×(sinx/cosx)
= 2
= -3(sin²x+cos²x)
= -3(1)
= -3
2. 2cosx over cos(90º-x) ×tanx
= 2cosx/sinx ×(sinx/cosx)
= 2
參考: me
2007-02-02 2:53 am
1) -1
2)=2(cosx over sinx)
=2 over tanx
2)=2(cosx over sinx)
=2 over tanx
收錄日期: 2021-04-13 00:31:42
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