We can offer two approaches to this problem - using trigonometry of right triangles (applicable for positive x) and purely trigonometric (applicable for all x, but we will use it for x<=0.
Let's analyze this problem from the position of trigonometry of the right triangle. For this, let's assume that 0<x<=1.
Then /_A=arcsin x is an angle from 0 to pi/2, sine of which is x.
Assume, this angle /_A is an acute angle of a right triangle Delta ABC. So, sin(/_A) is a ratio of an opposite cathetus a to hypotenuse c:
x=sin(/_A)=a/c
Consider another acute angle of this triangle - /_B.
Expression a/c represents its cosine.
Therefore,
/_B = arccos(a/c)=arccos x
Now we see that /_A=arcsin x and /_B=arccose x are two acute angles in a right triangle. Their sum is always pi/2.
Therefore,
sin(arcsin x+arccos x)=sin(pi/2)=1
Case with non-positive x we will consider differently.
If x <= 0, arcsin x is between -pi/2 and 0.
If x <= 0, arccos x is between pi/2 and pi.
Using formula for sin of a sum of two angles,
sin(arcsin x+arccos x) =
= sin(arcsin x)*cos(arccosx) + cos(arcsin x)*sin(arccos y)
By definition of inverse trigonometric functions arcsin and arccos, we write the following:
sin(arcsin x)=x
cos(arccos x)=x
Using trigonometric identity sin^2(phi)+cos^2(phi)=1, we can find the other components:
cos^2(arcsin x) = 1-sin^2(arcsinx) = 1-x^2
sin^2(arccos x) = 1-cos^2(arccos x) = 1-x^2
Since arcsin x is between -pi/2 and 0, its cosin is positive:
cos^2(arcsin x) = 1-x^2
=>cos(arcsin x) sqrt(1-x^2)
Since arccos x is between pi/2 and pi, its sine is positive:
sin^2(arccos x) = 1-x^2
=>sin(arccos x) sqrt(1-x^2)
Now we can calculate the value of the original expression:
sin(arcsin x+arccos x) =
= sin(arcsin x)*cos(arccosx) + cos(arcsin x)*sin(arccos y) =
= x*x+sqrt(1-x^2)*sqrt(1-x^2) =
= x^2 +1 -x^2 = 1
(as in the case we did geometrically).
