{"id":123037,"date":"2022-05-30T13:02:56","date_gmt":"2022-05-30T18:02:56","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=123037"},"modified":"2026-03-28T11:53:56","modified_gmt":"2026-03-28T16:53:56","slug":"half-angle-double-angle-and-product-trig-identities","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/half-angle-double-angle-and-product-trig-identities\/","title":{"rendered":"Half-Angle, Double Angle, and Product Trig Identities"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_426QTGct7RU\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_426QTGct7RU\" data-source-videoID=\"426QTGct7RU\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Half-Angle, Double Angle, and Product Trig Identities Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Half-Angle, Double Angle, and Product Trig Identities\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_426QTGct7RU:hover {cursor:pointer;} img#videoThumbnailImage_426QTGct7RU {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1741-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_426QTGct7RU\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_426QTGct7RU\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Half-Angle, Double Angle, and Product Trig Identities\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_426QTGct7RU\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_426QTGct7RU\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_426QTGct7RU\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction oFV_Function() {\n  var x = document.getElementById(\"oFV\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"oFV_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"oFV\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#HalfAngle_Identities\" class=\"smooth-scroll\">Half-Angle Identities<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Double_Angle_Identities\" class=\"smooth-scroll\">Double Angle Identities<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Product_Formulas\" class=\"smooth-scroll\">Product Formulas<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hello, and welcome to this video over some of the common trigonometric identities. In this video, we will go over the half-angle identities, double angle identities, and the product properties.<\/p>\n<h2><span id=\"HalfAngle_Identities\" class=\"m-toc-anchor\"><\/span>Half-Angle Identities<\/h2>\n<h3><span id=\"Finding_Sine_of_15_Degrees\" class=\"m-toc-anchor\"><\/span>Finding Sine of 15 Degrees<\/h3>\n<p>\nLet\u2019s jump right in by finding the sine of 15 degrees using the half-angle identities. Here\u2019s the half-angle identity for sine:<\/p>\n<div class=\"examplesentence\">\\(\\sin{(\\frac{\\theta}{2})}=\\pm \\sqrt{\\frac{1-\\cos{\\theta}}{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis is used to find the sine of half a known angle. For our problem, we can use it to find half of 30 degrees, which is 15 degrees. We could also use it to find half of 45 degrees, which is 22.5 degrees.<\/p>\n<p>To find our sine of 15 degrees, we\u2019ll use 30 degrees as \\(\\theta\\), since half of that will give us 15.<\/p>\n<p>As always, we start by gathering the information that we need and defining \\(\\theta\\).<\/p>\n<div class=\"examplesentence\">\\(\\theta=30\u00b0\\)<br \/>\n\\((\\frac{\\sqrt{3}}{2},\\frac{1}{2})\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we simply substitute all of the appropriate values into the formula.<\/p>\n<div class=\"examplesentence\">\\(\\sin{(\\frac{\\theta}{2})}= \\sqrt{\\frac{1-\\cos{(\\theta)}}{2}}\\)<br \/>\n\\(\\sin{(\\frac{30\u00b0}{2})}=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, we simplify.<\/p>\n<div class=\"examplesentence\">\\(\\sin{(15\u00b0)}=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{\\sqrt{2}}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis time we have a radical in our denominator so we need to rationalize it. The process for doing so is a bit different though since the radical is a monomial. We multiply the top and bottom of our answer by the denominator in order to make our new denominator an integer.<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{1-\\frac{\\sqrt{3}}{2}}}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}}\\) \\(=\\frac{\\sqrt{2}\\sqrt{1-\\frac{\\sqrt{3}}{2}}}{2}\\) \\(=\\frac{\\sqrt{2-\\sqrt{3}}}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd just like that, we have our answer.<\/p>\n<h3><span id=\"Cosine_and_Tangent\" class=\"m-toc-anchor\"><\/span>Cosine and Tangent<\/h3>\n<p>\nAs it turns out, we have half-angle identities for cosine and tangent as well. In fact, tangent has two!<\/p>\n<div class=\"examplesentence\">\\(\\cos{(\\frac{\\theta}{2})}=\\pm \\sqrt{\\frac{1+\\cos{\\theta}}{2}}\\)<\/p>\n<hr style=\"margin-top: 0; margin-bottom: 0.5em;\">\n\\(\\tan{(\\frac{\\theta}{2})}=\\frac{\\sin{\\theta}}{1+\\cos{\\theta}}\\)<br \/>\n\\(\\tan{(\\frac{\\theta}{2})}=\\frac{1-\\cos{\\theta}}{\\sin{\\theta}}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Finding_Tangent_of_22_5_Degrees\" class=\"m-toc-anchor\"><\/span>Finding Tangent of 22.5 Degrees<\/h3>\n<p>\nLet\u2019s go ahead and use one of the tangent half-angle identities to find the tangent of 22.5 degrees. Since 22.5 is half of 45, we can use 45 for \\(\\theta\\). It doesn\u2019t matter which of the tangent half-angle identities we choose; we\u2019ll get the same answer either way. Let\u2019s go with the first one.<\/p>\n<p>Once again, our process is the same. Write down the information we\u2019re going to need and then plug that information into our formula. <\/p>\n<div class=\"examplesentence\">\\(\\theta=45\u00b0\\)<br \/>\n\\((\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2})\\)<br \/>\n\\(\\tan{(\\frac{\\theta}{2})}=\\frac{\\sin{\\theta}}{1+\\cos{\\theta}}\\)<br \/>\n\\(\\tan{(\\frac{45\u00b0}{2})}=\\frac{\\frac{\\sqrt{2}}{2}}{1+\\frac{\\sqrt{2}}{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt gets a little tricky since we have fractions inside of fractions, but writing it out as fraction division makes that easier. In order to do this, we need to write the denominator as one fraction, which we accomplish by turning the one into a fraction \\((\\frac{2}{2})\\) and then rewriting the whole denominator. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{\\frac{\\sqrt{2}}{2}}{1+\\frac{\\sqrt{2}}{2}}=\\frac{\\frac{\\sqrt{2}}{2}}{\\frac{2+\\sqrt{2}}{2}}\\) \\(=\\frac{\\sqrt{2}}{2}\\div \\frac{2+\\sqrt{2}}{2}\\) \\(= \\frac{\\sqrt{2}}{2}\\cdot \\frac{2}{2+\\sqrt{2}}=\\frac{\\sqrt{2}}{2+\\sqrt{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, we have to rationalize our denominators by multiplying the top and bottom of our answers by the conjugate of the denominator, which looks just like the denominator but has the opposite sign between the two terms. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{\\sqrt{2}}{2+}\\cdot \\frac{2-\\sqrt{2}}{2-\\sqrt{2}}=\\frac{2\\sqrt{2-}\\sqrt{4}}{4-2\\sqrt{2}+2\\sqrt{2}-\\sqrt{4}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we can simplify even further and see if we can make the answers look the same.<\/p>\n<div class=\"examplesentence\">\\(\\frac{2\\sqrt{2}-2}{2}=\\sqrt{2}-1\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h2><span id=\"Double_Angle_Identities\" class=\"m-toc-anchor\"><\/span>Double Angle Identities<\/h2>\n<p>\nNext up are the double angle identities.<\/p>\n<div class=\"examplesentence\">\\(\\sin{(2\\theta)}=2 \\sin{\\theta} \\cos{\\theta}\\)<br \/>\n\\(\\cos{(2\\theta)}=\\cos^{2}{\\theta}-\\sin^{2}{\\theta}\\) <br \/>\n\\(\\tan{(2\\theta)}=\\frac{2 \\tan{\\theta}}{1-\\tan^{2}{\\theta}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe double angle identities aren\u2019t quite as useful when dealing with unit circle angles, since we already know the values for double the angles given on it. But they could definitely be used for a problem where a different angle was given (say 10 degrees) and we needed to find the cosine of 20 degrees.<\/p>\n<p>Or if we wanted to check to make sure the answer we found when using a half-angle formula was correct, the double angle formula will \u201cundo\u201d what we did with the half-angle formula. <\/p>\n<h3><span id=\"Finding_Cosine_of_120_Degrees\" class=\"m-toc-anchor\"><\/span>Finding Cosine of 120 Degrees<\/h3>\n<p>\nTo see how these work, let\u2019s find the cosine of 120 degrees by using the cosine double-angle identity. That means \\(\\theta\\) will be 60 degrees, which is on our unit circle. 120 degrees is also on our unit circle, so we will know quickly if our answer is correct!<\/p>\n<p>As usual, we set up the information we need and then write down our formula.<\/p>\n<div class=\"examplesentence\">\\(\\theta=60\u00b0\\)<br \/>\n\\((\\frac{1}{2},\\frac{\\sqrt{3}}{2})\\)<br \/>\n\\(\\cos{(2\\theta)}=\\cos^{2}{\\theta}-\\sin^{2}{\\theta}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, we substitute in our values. <\/p>\n<div class=\"examplesentence\">\\(\\cos{(2(60\u00b0))}=(\\frac{1}{2})^{2}-(\\frac{\\sqrt{3}}{2})^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWith the cosine double-angle formula, our substituted values are squared, which very nicely removes our radicals. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 120%;\">\\(\\frac{1}{4}-\\frac{3}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, we subtract our fractions and get our answer of \\(-\\frac{1}{2}\\).<\/p>\n<p>Checking the unit circle, we see that the cosine of 120 degrees is \\(-\\frac{1}{2}\\), so we\u2019re right! This is actually one of those cases where the calculator will give us the exact value too, since it\u2019s a rational answer. Also, note that the cosine of 120 degrees is negative. This is because 120 degrees is in quadrant II, where cosine is always negative. <\/p>\n<h2><span id=\"Product_Formulas\" class=\"m-toc-anchor\"><\/span>Product Formulas<\/h2>\n<p>\nOur last identities to cover are the formulas to find the product of the sines or cosines of two angles. For instance, what if we wanted to know the exact value of the sine of 45 degrees times the cosine of 15 degrees? We could do this by finding the cosine of 15 degrees using the half-angle cosine identity and then multiplying by the sine of 45 degrees from the unit circle. But there is another way, using the product formulas, which look like this: <\/p>\n<p>\\begin{align*}<br \/>\n\\cos \\alpha \\cos \\beta &#038;= \\frac{1}{2} [\\cos(\\alpha + \\beta) + \\cos(\\alpha &#8211; \\beta)] \\\\[1em]<br \/>\n\\sin \\alpha \\sin \\beta &#038;= \\frac{1}{2} [\\cos(\\alpha &#8211; \\beta) &#8211; \\cos(\\alpha + \\beta)] \\\\[1em]<br \/>\n\\sin \\alpha \\cos \\beta &#038;= \\frac{1}{2} [\\sin(\\alpha + \\beta) + \\sin(\\alpha &#8211; \\beta)] \\\\[1em]<br \/>\n\\cos \\alpha \\sin \\beta &#038;= \\frac{1}{2} [\\sin(\\alpha + \\beta) &#8211; \\sin(\\alpha &#8211; \\beta)]<br \/>\n\\end{align*}<\/p>\n<p>Notice that each one covers a different product. The first one is when we\u2019re finding the product of two cosines. The second when we\u2019re finding the product of two sines. The third when the product is of a sine and a cosine, and the last when it\u2019s a product of a cosine and a sine.<\/p>\n<p>Our problem is looking for \\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\), so we want to use the third one. <\/p>\n<p>We start by establishing \\(\\alpha\\) as 45 degrees and \\(\\beta\\) as 15 degrees. <\/p>\n<div class=\"examplesentence\">\\(\\alpha=45\u00b0\\)<br \/>\n\\(\\beta=15\u00b0\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we write down our formula, making sure we have all the signs right. <\/p>\n<div class=\"examplesentence\">\\(\\sin{(\\alpha)}\\cos{(\\beta)}\\)\\(\\)\\(=\\frac{1}{2}[\\sin{(\\alpha + \\beta)}+ \\sin{(\\alpha &#8211; \\beta)}]\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we substitute in our values for \\(\\alpha\\) and \\(\\beta\\). <\/p>\n<div class=\"examplesentence\">\\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\)\\(=\\frac{1}{2}[\\sin{(45\u00b0+15\u00b0)}\\)\\(+\\sin{(45\u00b0-15\u00b0)}]\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, we add and subtract our angles in the parentheses to find what we\u2019re going to need.<\/p>\n<div class=\"examplesentence\">\\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\)\\(=\\frac{1}{2}[\\sin{(60\u00b0)}+\\sin{(30\u00b0)}]\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe will need to know the sine of 60 degrees and the sine of 30 degrees. Heading back to the unit circle, we find those two values and plug them into our formula. <\/p>\n<div class=\"examplesentence\">\\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\)\\(=\\frac{1}{2}[\\frac{\\sqrt{3}}{2}+\\frac{1}{2}]\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, we simplify.<\/p>\n<div class=\"examplesentence\">\\(\\frac{1}{2}[\\frac{\\sqrt{3}+1}{2}]=\\frac{\\sqrt{3}+1}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can check our answer on a calculator by evaluating our answer and then finding out the approximate value of \\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\). In both cases, we get this:<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{3}+1}{4}=0.683012701892\\)<br \/>\n\\(\\sin{(45\u00b0)}\\cos{(15\u00b0)}\\)\\(=0.683012701892\\)<\/div>\n<p>\n&nbsp;<br \/>\nThese formulas can also be used to turn products into sums, or sums into products, especially when angles are given as variables. <\/p>\n<p>Hopefully, this review was helpful and you\u2019re well on your way to mastering these trig formulas. 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