Thread:

1. i identity

Demonstrate that $\arccos^i(\cosh1) = e^{-\frac{\pi}{2}}$

I am not a mathematician so sorry if this is not very challenging. Basically I just found this identity to be interesting and wonder how obvious/obscure it is to real mathematicians.

2. Note that $cosh(1)=cos(i)$

$cos^{-1}(cos(i))=i$

Therefore, we have $i^{i}$

Now, show this equals $e^{\frac{-\pi}{2}}$

3. Originally Posted by galactus
Note that $cosh(1)=cos(i)$

$cos^{-1}(cos(i))=i$

Therefore, we have $i^{i}$

Now, show this equals $e^{\frac{-\pi}{2}}$

Yep.

Just to tidy up a bit:

$\cos^{i}(\cosh1)^{-1}=i^i=e^{\frac{-\pi}{2}}$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8830580711364746, "perplexity_flag": "middle"}