1. cos(90°－A)分之cos2次A≡sin(90°－A)tan(90°－A)
2. Given that sin﹫=25分之7,find the value of 3tan﹫－cos﹫分之1by using Pythagoras' theorem.
maths1``問!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!help.......
2008-05-14 4:37 am
回答 (2)
2008-05-14 5:10 am
✔ 最佳答案
1. (cosA)^2/(cos(90°-A))= (cosA)(cosA)/(sinA)
= (cosA)(cosA)/(cosA)(tanA)
= (cosA)/(tanA)
= cosA tan(90°－A)
= sin(90°－A)tan(90°－A)
2. sin@=7/25
Consider a right-angled triangle with hypotenuse 25 and another side 7
the third side = root(25^2-7^2) -------(Pyth. theorem)
= root(576)
= 24
Then tan@= 7/24 and cos@= 24/25
3tan@-1/cos@
= 3x7/24-1/(24/25)
= 21/24-25/24
= -4/24
= -1/6
參考: Calculation
2008-05-14 5:10 am
1.
LHS
= cos2A/sinA
=cotAcosA
=sin(90-A)tan(90-A)=RHS
2.
sin@= 7/25 cos@ = 24/25 tan@ = 7/24
3(7/24)-24/25
= -17/200
LHS
= cos2A/sinA
=cotAcosA
=sin(90-A)tan(90-A)=RHS
2.
sin@= 7/25 cos@ = 24/25 tan@ = 7/24
3(7/24)-24/25
= -17/200
收錄日期: 2021-04-13 15:34:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080513000051KK02524