The first thing we need to consider is our additional formulae for trig:
#cos(A + B ) = cosAcosB - sinAsinB #
So for this problem we see that #color(red)(A = arccosx # and #color(blue)(B = arcsinx #
#=> cos( color(red)(arccosx) + color(blue)(arcsinx )) = #
#cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) ) #
We know:
#cos(arccosx) = x " and " sin(arcsinx) = x #
Now to find #cos(arcsinx) " and " sin(arccosx ) :#
#"let " color(orange)(theta_1 = arcsinx #
#=> sin theta_1 = x #
Use #color(green)( cos^2 x + sin^2 -= 1 #
#costheta_1 = sqrt(1-sin^2 theta_1 ) = sqrt(1-x^2 ) #
#=> cos(theta_1) = cos(arcsinx ) = sqrt(1-x^2 ) #
# "let " color(orange)(theta_2 = arccosx #
#=> costheta_2 = x #
#=> sintheta_2 = sqrt(1-x^2 ) #
#=> sin(arccosx) = sqrt(1-x^2 ) #
Hence this:
#cos( color(red)(arccosx) ) * cos color(blue)((arcsinx )) - sin color(red)( (arccosx) ) * sin color(blue)( (arcsinx) ) #
Becomes:
# xsqrt(1-x^2) - xsqrt(1-x^2) = color(red)(0 #
Note: This only holds for #|x| <= 1 #
As that is values of #x# that #arcsinx # and #arccosx# are valid for...