We first have to find in 1/4 the limited development of Arcsin(x)
So, we have to find LD_1(1/4) sin^(-1)(x)
Let X=x-1/4
We have to find LD(0) sin^(-1)(X+1/4)
also, f'(X)=1/sqrt(1-(X+1/4)^2)=1/sqrt(1-X^2-1/2X-1/16)
=1/sqrt(15/16-(X^2+1/2X))=16/15*1/sqrt(1-15/16(X^2+1/2X))
Let Y=-15/16(X^2+1/2X)
We have f'(Y)=16/15*1/sqrt(1+Y)=16/15*(1+Y)^(-1/2)
And we know a limited development of (1+Y)^n
We will only keep term of degree 0 and 1.
LD_1(0) (1+Y)^n=1+nY+o(Y)
Let replace Y by -15/16(X^2+1/2X) and n=-1/2
LD_1(0) f'(X)=1+15/32(color(red)(cancel(color(black)(X^2)))+1/2X)+o(X)
=1+15/64X
And LD_(n+1)(0) f(X)=f(0)+intLD_n(0)f(X)dX + o(X^(n+1))
<=>LD_2(0)sin^(-1)(X)=color(red)(cancel(color(black)(sin^(-1)(0))))^(=0)+int(1+15/64X)dX+o(X^2)
LD_2(0)sin^(-1)(X)=X+15/128X^2+o(X^2)
Then, finally, replace X=x-1/4, knowing that
we only want a LD_1(1/4)f(x)
LD_2(1/4)sin^(-1)(x)=x-1/4+15/128(x-1/4)^2+o((x-1/4)^2)
LD_1(1/4)sin^(-1)(x)=x-1/4+color(red)(cancel(color(black)(15/128x^2)))-15/256x+15/4096+o(x)
=241/256x-1009/4096
So, the local linearization of sin^(-1)(x) in a=1/4 is
y=241/256x-1009/4096