{"id":737,"date":"2016-07-18T17:54:00","date_gmt":"2016-07-18T22:54:00","guid":{"rendered":"https:\/\/trigonography.com\/?p=737"},"modified":"2022-06-25T01:42:04","modified_gmt":"2022-06-25T06:42:04","slug":"exponential-forms-of-hyperbolic-sine-and-cosine","status":"publish","type":"post","link":"https:\/\/trigonography.com\/?p=737","title":{"rendered":"Exponential Forms of Hyperbolic Sine and Cosine"},"content":{"rendered":"
\n
\"trigonograph-hyperbolicsincos\"<\/a><\/figure>\n<\/div>\n\n\n
$$\\begin{align}
2\\,\\sinh u &\\;=\\; e^{u} -\\, e^{-u} \\\\[4pt]
2\\,\\cosh u &\\;=\\; e^{u} + e^{-u}
\\end{align}$$<\/div>\n\n\n\n

$$\\begin{align}
|\\overline{OX}|\\cdot|\\overline{XY}| \\;\\equiv\\; 1 &\\quad\\to\\quad u = \\int_{1}^{|\\overline{OX}|} \\frac{1}{t}dt \\;=\\; \\ln |\\overline{OX}| \\\\[6pt]
&\\quad\\to\\quad |\\overline{OX}| = e^{u} \\quad\\text{and}\\quad |\\overline{XY}| = e^{-u}
\\end{align}$$<\/p>\n\n\n\n

\"trigonograph-hyperbolicsincos\"<\/a>\n

 <\/p>\n

Some background.<\/b> Geometrically, we define \\(\\sinh u\\) and \\(\\cosh u\\) by direct analogy with \\(\\sin\\theta\\) and \\(\\cos\\theta\\): as certain perpendicular segments associated with an arc of the “unit hyperbola”, \\(x^2 – y^2 = 1\\).<\/p>\n

While \\(\\theta\\) is usually interpreted as the length of a circular arc, we note that it is also twice<\/em> the area of the corresponding circular sector. The hyperbolic parameter \\(u\\) is interpreted via area, as well; today’s trigonograph shows why:<\/p>

<\/p>\n

Conveniently scaling lengths in the unit hyperbola figure by \\(\\sqrt{2}\\) —and, thus, scaling areas by \\(2\\)— we see that \\(\\sinh u\\) and \\(\\cosh u\\) are directly computable from \\(u\\) via the exponential function!<\/em>

(In fact, we can say the same of \\(\\sin\\theta\\) and \\(\\cos\\theta\\), but we need complex<\/em> exponentials for that.)<\/p>\n<\/div>\n\n\n\n

Taken from this answer of mine<\/a> on the Mathematics Stack Exchange.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

$$\\begin{align}2\\,\\sinh u &\\;=\\; e^{u} -\\, e^{-u} \\\\[4pt]2\\,\\cosh u &\\;=\\; e^{u} + e^{-u}\\end{align}$$ $$\\begin{align}|\\overline{OX}|\\cdot|\\overline{XY}| \\;\\equiv\\; 1 &\\quad\\to\\quad u = \\int_{1}^{|\\overline{OX}|} \\frac{1}{t}dt \\;=\\; \\ln |\\overline{OX}| \\\\[6pt]&\\quad\\to\\quad |\\overline{OX}| = e^{u} \\quad\\text{and}\\quad |\\overline{XY}| = e^{-u}\\end{align}$$   Some background. Geometrically, we define \\(\\sinh u\\) and \\(\\cosh u\\) by direct analogy with \\(\\sin\\theta\\) and \\(\\cos\\theta\\): as certain perpendicular segments associated with […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-737","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/737","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=737"}],"version-history":[{"count":10,"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/737\/revisions"}],"predecessor-version":[{"id":1197,"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/737\/revisions\/1197"}],"wp:attachment":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=737"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=737"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}