y= a + cosbx
(0,3) (2pi, 1)
I don't see how b is 1/2?
find a and b?
2016-08-19 5:09 pm
回答 (3)
2016-08-20 7:58 am
Using x = 0 and y = 3 we have:
3 = a + cos(0)
i.e. 3 = a + 1
=> a = 2
Hence, y = 2 + cosbx
With x = 2π and y = 1 we have:
1 = 2 + cos(2bπ)
i.e. cos(2bπ) = -1
=> 2bπ = π
so, 2b = 1
Hence, b = 1/2
Thus, y = 2 + cos(x/2)
:)>
3 = a + cos(0)
i.e. 3 = a + 1
=> a = 2
Hence, y = 2 + cosbx
With x = 2π and y = 1 we have:
1 = 2 + cos(2bπ)
i.e. cos(2bπ) = -1
=> 2bπ = π
so, 2b = 1
Hence, b = 1/2
Thus, y = 2 + cos(x/2)
:)>
2016-08-19 5:21 pm
3 = a + cos(0 * b)
1 = a + cos(2pi * b)
3 = a + cos(0)
3 = a + 1
2 = a
1 = 2 + cos(2pi * b)
-1 = cos(2pi * b)
cos(pi + 2pi * k) = cos(2pi * b)
pi + 2pi * k = 2pi * b
1 + 2k = 2b
1/2 + k = b
k is an integer
1 = a + cos(2pi * b)
3 = a + cos(0)
3 = a + 1
2 = a
1 = 2 + cos(2pi * b)
-1 = cos(2pi * b)
cos(pi + 2pi * k) = cos(2pi * b)
pi + 2pi * k = 2pi * b
1 + 2k = 2b
1/2 + k = b
k is an integer
2016-08-19 5:12 pm
y = a + cos(bx)
3 = a + cos(b*0)
3 = a + cos(0)
3 = a + 1
a = 3 - 1
a = 2
y = 2 + cos(bx)
1 = 2 + cos(b*pi)
cos(b*pi) = 1 - 2
cos(b*pi) = -1
b = any odd number
b can't be 1/2
3 = a + cos(b*0)
3 = a + cos(0)
3 = a + 1
a = 3 - 1
a = 2
y = 2 + cos(bx)
1 = 2 + cos(b*pi)
cos(b*pi) = 1 - 2
cos(b*pi) = -1
b = any odd number
b can't be 1/2
收錄日期: 2021-04-21 19:39:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160819090911AA4iIH9