Hi. I need to prove the following identity(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} } [/tex]

I was given a hint to write

[tex]\cos{A}=z[/tex],

then rewrite

[tex]\cos{A}[/tex]

in terms of the exponential.

[tex]\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z[/tex]

I took the log on both sides and got stuck at that point.

[tex] \ln{\exp{iA} + \exp{-iA}}=\ln{z^2} [/tex]

I know it's a correct method becaues the right hand is starting to take form. But i just couldn't solve for A (which will be [tex]\arccos{z}[/tex] right?).

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# Trigonometric identity

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