Is tanh−1(sin⁡2(x+dπ4))=1πln⁡(cot2x)? As you see the above graph that arctan⁡h(sin⁡2(x+π4)) is similar to 1πln⁡(cot2x) Does this mean that by some minor alterations in the functions, they would be equal? Why are both of functions so similar?

We have artanh(sin⁡2(x+π4))=artanh(cos⁡2x)=12ln⁡(1+cos⁡2x1−cos⁡2x) =12ln⁡(1+cos2x−sin2x1−cos2x+sin2x) =12ln⁡(2cos2x2sin2x) =12ln⁡(cot2x)

We have an answer to "Is tanh−1(sin⁡2(x+dπ4))=1πln⁡(cot2x)? As you see the above graph..."

Octavio Miller

2022-01-23

{tanh}^{- 1} (\sin 2 (x + d \frac{π}{4})) = \frac{1}{π} \ln (\cot^{2} x)

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As you see the above graph that

\arctan h (\sin 2 (x + \frac{π}{4}))

is similar to

\frac{1}{π} \ln (\cot^{2} x)

Does this mean that by some minor alterations in the functions, they would be equal?
Why are both of functions so similar?

Aaron Hughes

Beginner2022-01-24Added 13 answers

We have

a r tanh (\sin 2 (x + \frac{π}{4})) = a r tanh (\cos 2 x) = \frac{1}{2} \ln (\frac{1 + \cos 2 x}{1 - \cos 2 x})

= \frac{1}{2} \ln (\frac{1 + \cos^{2} x - \sin^{2} x}{1 - \cos^{2} x + \sin^{2} x})

= \frac{1}{2} \ln (\frac{2 \cos^{2} x}{2 \sin^{2} x})

= \frac{1}{2} \ln (\cot^{2} x)

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