Let f(x) = arcsin(x) + arccos(x)
We first prove that f(x) is a constant function. First find the
derivative of f.
f '(x) = d( arcsin(x) )/dx + d( arccos(x) )/dx
= 1 / sqrt(1  x^{2}) + (1 / sqrt(1  x^{2}) )
= 0
Now if f '(x) = 0 for all values of x, then that means that f(x) is a constant function that may be calculated using any value of x. Let us use x = 0 and x = 1.(Note that one value is enough).
f(0) = arcsin(0) + arccos(0) = 0 +π/2 =π/2
f(1) = arcsin(1) + arccos(1) =π/2 + 0 =π/2
Hence arcsin(x) + arccos(x) =π/2 for all values of x.
More on applications of differentiation
applications of differentiation
