prove a identityarcsin x+ arccos x=pi/2 -1=<x=<1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may use the following alternative method, such that:

`(arcsin x+ arccos x)' = (pi/2)' => 1/(sqrt(1 - x^2)) - 1/(sqrt(1 - x^2)) = 0 => 0 = 0`

Hence, if derivative of the function y` = arcsin x+ arccos x` is zero, thus, the function is constant.

Once proven that the function `y = arcsin x+ arccos x` is a constant, you only need to prove that the constant is `pi/2` .

You need to evaluate the function at several values of `x in [-1,1]` , such that:

`x = -1 => arcsin (-1)+ arccos (-1) = -pi/2 + pi = pi/2`

`x = 0 => arcsin 0 + arccos 0 = 0 + pi/2 = pi/2`

`x = 1 => arcsin 1 + arccos 1 = pi/2 + 0 = pi/2`

Hence, testing if the expression `arcsin x+ arccos x = pi/2` if `x in [-pi/2,pi/2]` , using the property of derivative of constant function, yields that `arcsin x+ arccos x = pi/2` is valid.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll take sine function both sides:

sin (arcsin x + arccos x)  =sin pi/2

Since sin pi/2 = 1, we'll replace it.

We'll apply the formula:

sin (a+b) = sin a*cos b + sin b*cos a

sin (arcsin x + arccos x) = sin(arcsin x)*cos(arccos x) + sin(arccos x)*cos (arcsin x)

We know that sin (arcsin t) = t and sin (arccos t) = sqrt(1 - t^2)

cos (arccos t) = t and cos(arcsin t) = sqrt(1 - t^2)

sin (arcsin x + arccos x) = x*x + sqrt[(1-x^2)^2]

sin (arcsin x + arccos x) = x^2 + 1 - x^2

We'll eliminate like terms:

sin (arcsin x + arccos x) = 1

Since LHS = RHS, then the identity arcsin x + arccos x  =  pi/2 is verified.

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