I need help.
I got a problem that I didn't know how to slove.
Here are my questions.
1 . Find a six trigonometric function of an given angle. If any are not defined say 'no defined '
-7π/6
2. Find the exact value of the expression. Do not use a calculator.
6cosπ/4 - 5 sec π/6
請高手指教
謝謝！
Help! Math ! 函數!
2014-10-03 7:25 pm
回答 (2)
2014-10-03 11:23 pm
✔ 最佳答案
1.sin(-7π/6)
= sin[2π + (-7π/6)]
= sin[π - (π/6)]
= sin[π/6]
= 1/2
cos(-7π/6)
= cos[π - (π/6)]
= -cos[π/6]
= -(√3)/2
tan(-7π/6)
= sin(-7π/6) / cos(-7π/6)
= (1/2) / [-(√3)/2]
= -1/√3
= -(√3)/3
sec(-7π/6)
= 1 / cos(-7π/6)
= 1 / [-(√3)/2]
= -2/√3
= -2(√3)/3
csc(-7π/6)
= 1 / sin(-7π/6)
= 1 / (1/2)
= 2
cot(-7π/6)
= 1 / tan(-7π/6)
= 1 / (-1/√3)
= -√3
====
2.
6cos(π/4) - 5sec(π/6)
= 6cos(π/4) - [5/cos(π/6)]
= 6[(√2)/2] - 5/[(√3)/2]
= (3√2) - (10/√3)
= (3√2) - [10(√3)/3]
= (9√2 - 10√3)/3
參考: 土扁
2014-10-03 9:17 pm
2.
cos(2*π/4)=0
2cos²(π/4)-1=0
cos²(π/4)=1/2
cos(π/4)=1/√2 or -1/√2 (rejected)
2014-10-03 13:19:47 補充：
cos(3π/6)=0
cos(π/6+2π/6)=0
cos(π/6)cos(2π/6)-sin(π/6)sin(2π/6)=0
cos(π/6)[2cos²(π/6)-1] - sin(π/6)[2sin(π/6)cos(π/6)]=0
2cos³(π/6)-cos(π/6) - 2[1-cos²(π/6)]cos(π/6)=0
4cos³(π/6) - 3cos(π/6)=0
cos(π/6)[4cos²(π/6)-3]=0
cos²(π/6)=3/4 or cos(π/6)=0(rejected)
cos(π/6)=(√3)/2 or -(√3)/2(rejected)
2014-10-03 13:20:02 補充：
∴cos(π/6)=(√3)/2 and cos(π/4)=1/√2
6cos(π/4) - 5 sec(π/6)
=6(1/√2) - 5/[(√3)/2]
=3√2 - (10√3)/3
=(9√2 - 10√3)/3
cos(2*π/4)=0
2cos²(π/4)-1=0
cos²(π/4)=1/2
cos(π/4)=1/√2 or -1/√2 (rejected)
2014-10-03 13:19:47 補充：
cos(3π/6)=0
cos(π/6+2π/6)=0
cos(π/6)cos(2π/6)-sin(π/6)sin(2π/6)=0
cos(π/6)[2cos²(π/6)-1] - sin(π/6)[2sin(π/6)cos(π/6)]=0
2cos³(π/6)-cos(π/6) - 2[1-cos²(π/6)]cos(π/6)=0
4cos³(π/6) - 3cos(π/6)=0
cos(π/6)[4cos²(π/6)-3]=0
cos²(π/6)=3/4 or cos(π/6)=0(rejected)
cos(π/6)=(√3)/2 or -(√3)/2(rejected)
2014-10-03 13:20:02 補充：
∴cos(π/6)=(√3)/2 and cos(π/4)=1/√2
6cos(π/4) - 5 sec(π/6)
=6(1/√2) - 5/[(√3)/2]
=3√2 - (10√3)/3
=(9√2 - 10√3)/3
收錄日期: 2021-04-15 16:42:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141003000051KK00020