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).
