prove arcsinx+arccosx = pi/2?

Question

where x exists in [-1,1]

how do i prove this?

thanks for the help

Answer 1

Let u = arcsin(x). Then
sin(u) = x

Let v = arccos(x). Then
cos(v) = x

It follows that
sin(u) = cos(v), so
sin(u) = sin(pi/2 - v), so
u = pi/2 - v, and
u + v = pi/2

But u = arcsin(x) and v = arccos(x), so

arcsin(x) + arccos(x) = pi/2

Answer 2

consider a right triangle ABO with hypoteneuse 1, and A = arcsin(x). By definition of sine, the side a (opposite A) has length x. By definition of cosine, the side a adjacent to B has length x. So B is arccos(x). The sum of the angles in the triangle = pi, and angle O = pi/2, so A + B = arcsin(x) + arccos(x) = pi/2

Answer 3

Draw a right angle triangle with angles a and b.
1 is the hyopetenuse.
x is opposite angle a
a + b = π/2
sin ^(-1) x + cos^(-1)x = π/2