I currently try to simplify the following trigonometric

markush35q

markush35q

Answered question

2022-03-29

I currently try to simplify the following trigonometric expression:
arcsin(2t1)+2arctan(1tt)

Answer & Explanation

Raiden Griffin

Raiden Griffin

Beginner2022-03-30Added 13 answers

Set
ft)=arcsin(2t1)+2arctan(1tt)
t(0,1]
by differentiating
f(t)=0, t(0,1]
thus f is constant on (0,1] putting t=1 you get f(1)=π2 giving
arcsin(2t1)+2arctan(1tt)=π2
t(0,1]
Ishaan Stout

Ishaan Stout

Beginner2022-03-31Added 14 answers

sin1(2t1)+2tan1(1tt), where 0<t1
Now using sin1(x)+cos1(x)=π2sin1(x)=π2cos1(x)
Now expression is π2cos1(2t1)+2tan1(1tt)
Now put t=cos2ϕ, then 0<cos2ϕ11cosϕ1[0]
So we get =π2cos1(2cos2ϕ1)+2tan1(1cos2ϕsin2ϕ)
So we get =π2cos1(cos2ϕ)+2tan1(tanϕ), where 0ϕπ[π2]

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