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Maths problem: arcsinx - arccosx = pi/2

Can someone point out what is wrong with this proof?

y = cos(x)
arccos(y) = x

sin(arccos(y) + pi/2 ) = cos(arccos(y)) -because sin(x+pi/2) = cos(x)

sin(arccos(y) + pi/2 ) = y

arccos(y) + pi/2 = arcsin(y)

arcsin(y)-arccos(y) = pi/2
(edited 6 years ago)
Original post by BeyondDoubt
Can someone point out what is wrong with this proof?

y = cos(x)
arccos(y) = x

sin(arccos(y) + pi/2 ) = cos(arccos(y)) -because sin(x+pi/2) = cos(x)

sin(arccos(y) + pi/2 ) = y

arccos(y) + pi/2 = arcsin(y)

arcsin(y)-arccos(x) = pi/2


the x should be a y ^^^
Original post by BeyondDoubt

sin(arccos(y) + pi/2 ) = y

arccos(y) + pi/2 = arcsin(y)


Here

arcsin⁡(sin⁡(x))=x\arcsin(\sin(x)) = x holds only in the interval [−π2,π2]\displaystyle \left[-\frac \pi 2, \frac \pi 2\right], and arccos⁡(y)+π2\displaystyle \arccos(y) + \frac \pi 2 has a range of [π2,3π2]\displaystyle \left[\frac \pi 2,\frac{3\pi} 2\right]. If you instead use π2−arccos⁡(y)\displaystyle \frac \pi 2 - \arccos(y), which has a range of [−π2,π2]\displaystyle\left[-\frac \pi 2, \frac \pi 2\right], you will get the correct result.
(edited 6 years ago)
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BeyondDoubt
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Original post by _gcx
Here

arcsin⁡(sin⁡(x))=x\arcsin(\sin(x)) = x holds only in the interval [−π2,π2]\displaystyle \left[-\frac \pi 2, \frac \pi 2\right], and arccos⁡(y)+π2\displaystyle \arccos(y) + \frac \pi 2 has a range of [π2,3π2]\displaystyle \left[\frac \pi 2,\frac{3\pi} 2\right]. If you instead use π2−arccos⁡(y)\displaystyle \frac \pi 2 - \arccos(y), which has a range of [−π2,π2]\displaystyle\left[-\frac \pi 2, \frac \pi 2\right], you will get the correct result.


Thank you!

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