Yet another inverse trig question (1 Viewer)

[Majishan]

the question is to derive arcsin(cos x)

i get an answer of (-sin x)/(root(1-cos^2 x))

but the answer its meant to be is one. little help

 

[pLuvia]

d/dx(arcsin[cosx])
=(-sinx)(1/sqrt{1-cos²x})
=-sinx/sqrt{1-cos²x}
=-sinx/sqrt{sin²x}
=-sinx/|sinx|

If sinx>0 then dy/dx=-1
If sinx<0 then dy/dx=1

 

[Majishan]

oh, now it seems easy. just needed to remember the old trig identities and things.
thanx