ArcSinx - ArcCosx =??? Option 1: pi/2 Option 2: -1 Option 3: pi/6?
回答 (3)
2020-04-29 12:02 pm
Let cos(θ) = x
Then arccos(x) = θ …… [1]
Trigonometric identity: cos(θ) = sin[(π/2) - θ]
Hence, sin[(π/2) - θ] = x
Then, arcsin(x) = (π/2) - θ …… [2]
[1] - [2]:
arcsin(x) - arccos(x)
= [(π/2) - θ] - θ
= (π/2) - 2θ
= (π/2) - 2arccos(x)
Hence, none of the three options is the answer.
The value of arcsin(x) - arccos(x) depends on the value of x.
Then arccos(x) = θ …… [1]
Trigonometric identity: cos(θ) = sin[(π/2) - θ]
Hence, sin[(π/2) - θ] = x
Then, arcsin(x) = (π/2) - θ …… [2]
[1] - [2]:
arcsin(x) - arccos(x)
= [(π/2) - θ] - θ
= (π/2) - 2θ
= (π/2) - 2arccos(x)
Hence, none of the three options is the answer.
The value of arcsin(x) - arccos(x) depends on the value of x.
2020-04-30 10:52 am
sin^(-1)(x) - cos^(-1)(x) = 1/sqrt2
2020-04-29 11:02 am
Depends on what x is.
收錄日期: 2021-04-18 18:34:30
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