y = (cos^2)x - sin x + 1
thanks
更新1:
唔係好明點解 -(sinx+1/2)^2+(1/4)+2 之後 = -(sinx+1/2)^2+(9/4) ?
trigon function...thanks
2007-01-27 11:27 pm
If x takes any real values , find the range of the values of y of the following function.
y = (cos^2)x - sin x + 1
thanks
y = (cos^2)x - sin x + 1
thanks
回答 (2)
2007-01-27 11:37 pm
✔ 最佳答案
y= (cos^2)x - sin x + 1
=(1-sin^2x)-sinx+1
=-(sin^2+sinx-2)
=-(sinx+1/2)^2+(1/4)+2
=-(sinx+1/2)^2+(1/4)+2
=-(sinx+1/2)^2+(9/4)
since x takes any real values
-1<=sinx<=1
when x takes -1/2 , y gets the maximum 9/4
when x takes 1, y gets the maximum -9/4+9/4=0
So
0<=y<=9/4
2007-01-27 15:38:33 補充：
應是 y gets the minimum -9/4+9/4=0
2007-01-27 21:00:59 補充：
(1/4)+2=(1/4)+(8/4)=9/4
2007-01-27 11:43 pm
y = (cos^2)x - sin x + 1
=[1-(sin x)^2]-sin x+1
=-(sin x)^2-sin x+2
=-(sin x+1/2)^2+9/4
so,0<9/4(because -1
2007-01-27 15:44:09 補充：
is
=[1-(sin x)^2]-sin x+1
=-(sin x)^2-sin x+2
=-(sin x+1/2)^2+9/4
so,0<9/4(because -1
2007-01-27 15:44:09 補充：
is
收錄日期: 2021-04-25 16:51:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070127000051KK02367