How to proof \(\displaystyle{\arcsin{{\left(\sqrt{{{2}\cdot{b}\cdot{\left(\sqrt{{{1}+{b}^{{2}}}}-{b}\right)}}}\right)}}}={\arccos{{\left(\sqrt{{{1}+{b}^{{2}}}}-{b}\right)}}}\)

Julian Peck

Julian Peck

Answered question

2022-04-06

How to proof
arcsin(2b(1+b2b))=arccos(1+b2b)

Answer & Explanation

WigwrannyErarmbmk

WigwrannyErarmbmk

Beginner2022-04-07Added 13 answers

Start from
(1+x2x)2=12x1+x2+x2
So we find the key relation
1(1+x2x)2=2x1+x2x2=2x(1+x2x)
Then use sin2+cos2=1 to say that
cos1(t)=sin1(1t2)
(true for all |t|1). If we use this and substitue t=1+x2x then the key relation says
cos1(1+x2x)=sin1(1(1+x2x)2)
=sin1(2x(1+x2x))
which is your relationship, with x replacing b.

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