F.4 A.Maths (the subsidiary angle)!

2006-10-27 5:03 am
Find the maximum & minimum value of the following expression;

1, (2cosX + 3sinX)^2

2, 1/(sinX - 2cosX)^2

回答 (2)

2006-10-27 5:13 am
✔ 最佳答案
LE = less than or equal to
GE = greater than or equal to

1. -sqrt(2^2+3^2) LE (2 cos X + 3 sin X) GE sqrt(2^2+3^2)
-sqrt(13) LE (2 cos X + 3 sin X) GE sqrt(13)
0 LE (2 cos X + 3 sin X)^2 GE 13
So, max = 13, min = 0

2. Same as above,
0 LE (sinX - 2cosX)^2 GE 5
1/(sinX - 2cosX)^2 LE 1/5
So, min = 1/5, no max value.
參考: My mathematical knowledge
2006-10-27 5:16 am
the principle is :

step 1 : develop the expression
step 2 : differentiate once and set the expression to zero
step 3 : then you can find out the max. & min. value of x

try by yourself la !!


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