y = arcsinx , -1 ≤ x ≤ 1
By expressing arccosx in terms of y, show that:
arcsinx + arccosx = pi/2
???
Absolutely no clue. Thanks in advance.
Help with arcsin question...?
2020-09-02 3:53 pm
回答 (2)
2020-09-02 5:22 pm
arcsin(x) = y …… [1]
Then sin(y) = x …… [2]
Trigonometric identity: cos[(π/2) - y] = sin(y) …… [3]
Plug [2] into [3]:
cos[(π/2) - y] = x
arccos(x) = (π/2) - y …… [4]
[1] + [4]:
arcsin(x) + arccos(x) = y + [(π/2) - y]
Hence, arcsin(x) + arccos(x) = π/2
Then sin(y) = x …… [2]
Trigonometric identity: cos[(π/2) - y] = sin(y) …… [3]
Plug [2] into [3]:
cos[(π/2) - y] = x
arccos(x) = (π/2) - y …… [4]
[1] + [4]:
arcsin(x) + arccos(x) = y + [(π/2) - y]
Hence, arcsin(x) + arccos(x) = π/2
2020-09-02 4:28 pm
arcsin(x) + arccos(x) = pi/2
cos(arcsin(x) + arccos(x)) = cos(pi/2)
cos(arcsin(x))cos(arccos(x)) - sin(arcsin(x))sin(arccos(x)) = 0
sqrt(1 - sin(arcsin(x))^2) * x - x * sqrt(1 - cos(arccos(x))^2) = 0
sqrt(1 - x^2) * x - x * sqrt(1 - x^2) = 0
0 = 0
cos(arcsin(x) + arccos(x)) = cos(pi/2)
cos(arcsin(x))cos(arccos(x)) - sin(arcsin(x))sin(arccos(x)) = 0
sqrt(1 - sin(arcsin(x))^2) * x - x * sqrt(1 - cos(arccos(x))^2) = 0
sqrt(1 - x^2) * x - x * sqrt(1 - x^2) = 0
0 = 0
收錄日期: 2021-04-24 08:02:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200902075310AAT3B1h