Is \(\displaystyle{\arccos{{\left({\frac{{\sqrt{{6}}+{1}}}{{{2}\sqrt{{3}}}}}\right)}}}={\arctan{{\left({\frac{{\sqrt{{3}}-\sqrt{{2}}}}{{{1}+\sqrt{{6}}}}}\right)}}}\)? They are equal according to the

Annie Rice

Annie Rice

Answered question

2022-04-16

Is arccos(6+123)=arctan(321+6)?
They are equal according to the calculator but how?
I made a triangle with base 6+1 and hypotenuse 23
Then the height comes out to be 526 using Pythagoras theorem.
But it should come out to be 32

Answer & Explanation

chabinka61jx

chabinka61jx

Beginner2022-04-17Added 12 answers

All you did is fine. Now, notice that
(32)2=3+226=526

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