Given #sin^-1 x +2 cos^-1 x = pi/2#
Therefore sin#(sin^-1 x +2 cos^-1x) = sin pi/2#
#sin (sin^-1 x) cos (2cos^-1 x) +cos (sin^-1 x) sin (2cos^-1 x) =1#
#sin(sin^-1 x)# equals x
#sin(cos^-1 x)# equals #sqrt(1-x^2)#
Now for evaluating #cos(2cos^-1 x)# and #sin(2cos^-1 x)#, let #cos^-1 x= theta#, so that #cos theta =x and sin theta = sqrt(1-x^2)#
#cos(2cos^-1 x) = cos 2theta= 2cos^2 theta -1= 2x^2 -1#
#sin(2cos^-1 x)= sin 2 theta= 2 cos theta sin theta =2xsqrt(1-x^2)#
Using these values,
#sin (sin^-1 x) cos (2cos^-1 x) +cos (sin^-1 x) sin (2cos^-1 x) =1#
would become
#x (2x^2 -1) +sqrt (1-x^2) 2x sqrt(1-x^2)# =1
#x(2x^2-1) +2x(1-x^2) =1#
which simplifies to x=1
The above result is although quite obvious from the text of the question itself, realising that #Sin^-1 1= pi/2# and #cos^-1 1 =0#
