Solve using analythic methods arcsinx + arccosx = pi/2 if x belongs to [ -1, 1]

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f(x) = arcsinx + arccosx = pi/2

If f(x) equals a constant number , then the derivative of f is zero. Let us verify:

f'(x) = (arcsinx + arccosx)'

      = 1/sqrt(1-x^2) - 1/sqrt(1-x^2) =0

Then f(x) = C   ( c is a constant number)

but f(1) = arcsin1 + arccos1...

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f(x) = arcsinx + arccosx = pi/2

If f(x) equals a constant number , then the derivative of f is zero. Let us verify:

f'(x) = (arcsinx + arccosx)'

      = 1/sqrt(1-x^2) - 1/sqrt(1-x^2) =0

Then f(x) = C   ( c is a constant number)

but f(1) = arcsin1 + arccos1 = pi/2 + 0= pi/2

Then C= pi/2

==> arcsinx + arccosx = pi/2

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