g(x) = -3sin(2π/15(x-3))+1
Find an equivalent function using cosine
Trig Functions?
2020-07-24 2:32 pm
回答 (2)
2020-07-24 8:55 pm
Somehow, I think you mean:
g(x) = -3sin[2π(x - 3)/15] + 1
Now, cosA = sin(π/2 - A)
Hence, g(x) = -3cos[(π/2) - 2π(x - 3)/15] + 1
Simplifying (π/2) - 2π(x - 3)/15, we get:
(15π/30) - 4π(x - 3)/30
i.e. (π/30)(15 - 4(x - 3)) => (π/30)(27 - 4x)
Also, cosA = cos(-A) so,
cos[(π/30)(27 - 4x)] => cos[(π/30)(4x - 27)]
Hence, g(x) = -3cos[(π/30)(4x - 27)] + 1
:)>
g(x) = -3sin[2π(x - 3)/15] + 1
Now, cosA = sin(π/2 - A)
Hence, g(x) = -3cos[(π/2) - 2π(x - 3)/15] + 1
Simplifying (π/2) - 2π(x - 3)/15, we get:
(15π/30) - 4π(x - 3)/30
i.e. (π/30)(15 - 4(x - 3)) => (π/30)(27 - 4x)
Also, cosA = cos(-A) so,
cos[(π/30)(27 - 4x)] => cos[(π/30)(4x - 27)]
Hence, g(x) = -3cos[(π/30)(4x - 27)] + 1
:)>
收錄日期: 2021-04-24 07:53:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200724063218AAVDc2K