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
