# 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 = pi/2 + 0= pi/2

Then C= pi/2

==> arcsinx + arccosx = pi/2

To solve arcsinx+arc cosx = pi/2. if x belongs to (-1,1).

Solution:

Let x = k and

Let -1 < k < 1. Then,

arcsink +arccosk = arc sink + arc sin (pi/2-k) = k+pi/2-k = pi/2. But to use the analytic method, we assume that ark sink +arccosk = f(k).

Differentiating we get

= 1/sqrt(1-k^2) - 1/sqrt(1-k) = f'(k). Or

)r = 0 = f'(k). So

arcsinx + arccosx = constant c.

When x=0, arc sin0 +arc cos0 = constant. Or

0 + pi/2 = const. So arc sinx +arc cosx = c = pi/2.

We'll note the expression (arcsin x + arccos x) as a function f(x).

For demonstrating that f(x)=pi/2 is a constant function, we have to differentiate the function f(x).

If this derivative is 0, that means that f(x)=pi/2, is a constant function, knowing the fact that a derivative of a constant function is 0.

f'(x) = (arcsin x + arccos x)'

f'(x) = 1/sqrt(1-x^2) - 1/sqrt(1-x^2)=0, so f(x)=constant

**f(1)=arcsin 1 + arccos 1 = pi/2 + 0=pi/2**