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    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
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    (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 ^^^
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    (Original post by BeyondDoubt)
    sin(arccos(y) + pi/2 ) = y

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

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

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