{"id":1146,"date":"2022-06-20T23:23:21","date_gmt":"2022-06-21T04:23:21","guid":{"rendered":"https:\/\/trigonography.com\/?p=1146"},"modified":"2022-06-20T23:23:21","modified_gmt":"2022-06-21T04:23:21","slug":"half-angle-identities-in-a-triangle","status":"publish","type":"post","link":"https:\/\/trigonography.com\/?p=1146","title":{"rendered":"Half-Angle Identities in a Triangle"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\" style=\"font-size:16px\">In his <a rel=\"noreferrer noopener\" href=\"https:\/\/geometriadominicana.blogspot.com\/2020\/06\/another-proof-for-two-well-known.html\" target=\"_blank\">Geometria Dominicana<\/a> blog, Emmanuel Jos\u00e9 Garc\u00eda has been championing a pair of half-angle identities<br> in a triangle. I thought I&#8217;d contribute to the cause by offering trigonographic proofs.<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><a href=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle-1024x737.png\" alt=\"\" class=\"wp-image-1149\" width=\"693\" height=\"499\" srcset=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle-1024x737.png 1024w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle-300x216.png 300w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle-768x553.png 768w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/cos-halfangle.png 1500w\" sizes=\"auto, (max-width: 693px) 100vw, 693px\" \/><\/a><\/figure>\n<\/div>\n\n\n<p class=\"has-text-align-center wp-block-paragraph\" style=\"font-size:16px\">$$\\begin{align}|CB&#8217;||CD|=|CQ||CQ&#8217;| &amp;\\quad\\to\\quad 2a\\cos\\frac12C\\cdot2b\\cos\\frac12C=(a+b+c)(a+b-c) \\\\[8pt] &amp;\\quad\\to\\quad \\cos^2\\frac12C=\\frac{s(s-c)}{ab} \\qquad\\qquad \\left(\\;s:=\\frac12(a+b+c)\\;\\right) \\end{align}$$<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><a href=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle-1024x922.png\" alt=\"\" class=\"wp-image-1150\" width=\"558\" height=\"501\" srcset=\"https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle-1024x922.png 1024w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle-300x270.png 300w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle-768x691.png 768w, https:\/\/trigonography.com\/blog\/wp-content\/uploads\/2022\/06\/sin-halfangle.png 1200w\" sizes=\"auto, (max-width: 558px) 100vw, 558px\" \/><\/a><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\" style=\"font-size:16px\">$$\\begin{align}|PB||PA&#8217;|=|PQ||PQ&#8217;| &amp;\\quad\\to\\quad 2a\\sin\\frac12C\\cdot2b\\sin\\frac12C=(-a+b+c)(a-b+c) \\\\[8pt] &amp;\\quad\\to\\quad \\sin^2\\frac12C=\\frac{(s-a)(s-b)}{ab}\\end{align}$$<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In his Geometria Dominicana blog, Emmanuel Jos\u00e9 Garc\u00eda has been championing a pair of half-angle identities in a triangle. I thought I&#8217;d contribute to the cause by offering trigonographic proofs. $$\\begin{align}|CB&#8217;||CD|=|CQ||CQ&#8217;| &amp;\\quad\\to\\quad 2a\\cos\\frac12C\\cdot2b\\cos\\frac12C=(a+b+c)(a+b-c) \\\\[8pt] &amp;\\quad\\to\\quad \\cos^2\\frac12C=\\frac{s(s-c)}{ab} \\qquad\\qquad \\left(\\;s:=\\frac12(a+b+c)\\;\\right) \\end{align}$$ $$\\begin{align}|PB||PA&#8217;|=|PQ||PQ&#8217;| &amp;\\quad\\to\\quad 2a\\sin\\frac12C\\cdot2b\\sin\\frac12C=(-a+b+c)(a-b+c) \\\\[8pt] &amp;\\quad\\to\\quad \\sin^2\\frac12C=\\frac{(s-a)(s-b)}{ab}\\end{align}$$<\/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-1146","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/1146","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=1146"}],"version-history":[{"count":10,"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/1146\/revisions"}],"predecessor-version":[{"id":1169,"href":"https:\/\/trigonography.com\/index.php?rest_route=\/wp\/v2\/posts\/1146\/revisions\/1169"}],"wp:attachment":[{"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1146"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1146"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/trigonography.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1146"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}