{"id":121192,"date":"2022-05-06T08:01:42","date_gmt":"2022-05-06T13:01:42","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=121192"},"modified":"2026-03-28T11:51:58","modified_gmt":"2026-03-28T16:51:58","slug":"sum-and-difference-trigonometric-identities","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/sum-and-difference-trigonometric-identities\/","title":{"rendered":"Sum and Difference Trigonometric Identities"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_QAyD412u4N0\" 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_QAyD412u4N0\" data-source-videoID=\"QAyD412u4N0\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Sum and Difference Trigonometric Identities Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Sum and Difference Trigonometric Identities\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_QAyD412u4N0:hover {cursor:pointer;} img#videoThumbnailImage_QAyD412u4N0 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1740-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_QAyD412u4N0\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_QAyD412u4N0\" 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\",\"Sum and Difference Trigonometric Identities\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_QAyD412u4N0\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_QAyD412u4N0\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_QAyD412u4N0\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction Dkz_Function() {\n  var x = document.getElementById(\"Dkz\");\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=\"Dkz_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=\"Dkz\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Unit_Circle\" class=\"smooth-scroll\">Unit Circle<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Sum_Identities\" class=\"smooth-scroll\">Sum Identities<\/a>\n<ul><\/li>\n<li class=\"toc-h3\"><a href=\"#Finding_the_Sine_of_the_Sum_of_Two_Angles\" class=\"smooth-scroll\">Finding the Sine of the Sum of Two Angles<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Cosine\" class=\"smooth-scroll\">Cosine<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Tangent\" class=\"smooth-scroll\">Tangent<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Simplifying_Answers\" class=\"smooth-scroll\">Simplifying Answers<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"toc-h2\"><a href=\"#Difference_Identities\" class=\"smooth-scroll\">Difference Identities<\/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>Hi, and welcome to this review of the sum and difference trigonometric identities!<\/p>\n<p>We have a lot of useful formulas to cover in this video, so let\u2019s get started!<\/p>\n<h2><span id=\"Unit_Circle\" class=\"m-toc-anchor\"><\/span>Unit Circle<\/h2>\n<p>\nBefore we dive in, let\u2019s take a moment to review the unit circle. The unit circle gives you the sine and cosine values for some of the most common angle measures.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-121204\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/05\/Sum-and-Difference-Trigonometric-Identities.jpg\" alt=\"the trigonometric unit circle\" width=\"652\" height=\"369\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/05\/Sum-and-Difference-Trigonometric-Identities.jpg 652w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/05\/Sum-and-Difference-Trigonometric-Identities-300x170.jpg 300w\" sizes=\"auto, (max-width: 652px) 100vw, 652px\" \/><\/p>\n<p>With this information, combined with some of the formulas we\u2019re going to cover today, we can find the exact trigonometric values of even more angles on the unit circle, all without using a calculator. <\/p>\n<h2><span id=\"Sum_Identities\" class=\"m-toc-anchor\"><\/span>Sum Identities<\/h2>\n<p>\nLet\u2019s start with the sum identities.<\/p>\n<h3><span id=\"Finding_the_Sine_of_the_Sum_of_Two_Angles\" class=\"m-toc-anchor\"><\/span>Finding the Sine of the Sum of Two Angles<\/h3>\n<p>\nThe first one we\u2019ll use is the formula for finding the sine of the sum of two angles.<\/p>\n<p>Let\u2019s look at the formula:<\/p>\n<div class=\"examplesentence\">\\(\\text{ sin}(a+\\beta )=\\text{ sin }\\alpha \\text{ cos }\\beta + \\text{ cos }\\alpha \\text{ sin }\\beta\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo if we want to find the sine of 75 degrees, we can do this by finding two angles that add up to 75 and using them as \\(\\alpha\\) and \\(\\beta\\). Fortunately, that\u2019s easy to find on our unit circle. Since \\(30+45=75\\), we can use 30 as \\(\\alpha\\) and 45 as \\(\\beta\\). We could also use 45 as \\(\\alpha\\) and 30 as \\(\\beta\\). It doesn&#8217;t really matter.<\/p>\n<p>First, let\u2019s write down the information we need for our two angles. We\u2019re using 30 degrees as \\(\\alpha\\), so we&#8217;re going to write down, \\(\\alpha\\)=30\u00b0, and then we&#8217;ll look up our cosine and our sine values from our unit circle. <\/p>\n<div class=\"examplesentence\">\\(\\alpha=30\u00b0\\)<br \/>\n\\(\\text{cos}(30\u00b0)=\\frac{\\sqrt{3}}{2}\\)<br \/>\n\\(\\text{sin}(30\u00b0)=\\frac{1}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we do the same thing for \\(\\beta\\), which we\u2019ve set as 45 degrees. <\/p>\n<div class=\"examplesentence\">\\(\\beta=45\u00b0\\)<br \/>\n\\(\\text{cos}(45\u00b0)=\\frac{\\sqrt{2}}{2}\\)<br \/>\n\\(\\text{sin}(45\u00b0)=\\frac{\\sqrt{2}}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we\u2019ll write down our formula for the sine sum identity.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(\\alpha +\\beta)=\\text{sin}(\\alpha)\\text{cos}(\\beta)+\\text{cos}(\\alpha)\\text{sin}(\\beta)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we\u2019ll substitute 30 degrees for \\(\\alpha\\), 45 degrees for \\(\\beta\\), and then substitute all of our sine and cosine values into the right side of our formula. <\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(30\u00b0+45\u00b0)=\\text{sin}(30\u00b0)\\text{cos}(45\u00b0)+\\text{cos}(30\u00b0)\\text{sin}(45\u00b0)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we&#8217;re going to substitute in our sine and cosine values.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(75\u00b0)=(\\frac{1}{2})(\\frac{\\sqrt{2}}{2})+(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{2}}{2})\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we&#8217;re going to simplify with a little bit of multiplication.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(75\u00b0)=\\frac{\\sqrt{2}}{4}+\\frac{\\sqrt{6}}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFinally, because both terms have the same denominator, we can write our answer as a single fraction.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(75\u00b0)=\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAs you can see, it\u2019s important to be strong in handling radicals and fractions to do this kind of work. We can check our answer by using a calculator. If we add the \\(\\sqrt{2}\\) to the \\(\\sqrt{6}\\) and then divide by 4, we get approximately 0.966. <\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(75\u00b0)=\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\approx 0.966\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we then use a calculator to find the sine of 75\u00b0, we get the same thing. <\/p>\n<p>But that begs the question, \u201cWhy don\u2019t we just use a calculator to find the sine instead of doing all this work?\u201d The answer is that the calculator is giving us an approximate value of the sine of 75 degrees. If we want to know the <em>exact value<\/em>, we need to do it the long way. When taking a standardized test, if you see the word \u201cexact\u201d in reference to a problem like this, then you\u2019ll know how to do this process.<\/p>\n<h3><span id=\"Cosine\" class=\"m-toc-anchor\"><\/span>Cosine<\/h3>\n<p>\nWhat if we want to find the <em>exact<\/em> value of the cosine of 75 degrees? We have a sum identity formula for that too:<\/p>\n<div class=\"examplesentence\">\\(\\text{cos}(\\alpha+\\beta)=\\text{cos}(\\alpha) \\text{cos}(\\beta)-\\text{ sin}(\\alpha) \\text{sin}(\\beta)\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt works in a similar way. We can plug in our values and find the cosine pretty quickly.<\/p>\n<p>We start by writing down our cosine sum identity. It\u2019s really easy to mix up the values if you don\u2019t do this, so it\u2019s always a good idea. Then, we\u2019ll substitute 30 degrees for \\(\\alpha\\), 45 degrees for \\(\\beta\\), and then substitute all of our sine and cosine values into the right side of the formula just like we did with the last problem. <\/p>\n<div class=\"examplesentence\">\\(\\text{cos}(30\u00b0+45\u00b0)=\\text{cos}(30\u00b0)\\text{cos}(45\u00b0)-\\text{ sin}(30\u00b0)\\text{sin}(45\u00b0)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we&#8217;ll plug in our sine and cosine values. So, first we&#8217;ll simply this left side.<\/p>\n<div class=\"examplesentence\">\\(\\text{cos}(75\u00b0)=(\\frac{\\sqrt{3}}{2})(\\frac{\\sqrt{2}}{2})-(\\frac{1}{2})(\\frac{\\sqrt{2}}{2})\\)<\/div>\n<p>\n&nbsp;<br \/>\nOnce again we do some multiplication and write our answer as a single fraction.<\/p>\n<div class=\"examplesentence\">\\(\\text{cos}(75\u00b0)=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nJust like the last problem, we can check our work by evaluating this radical expression on a calculator and then finding the approximate value of the cosine of 75\u00b0 on the calculator. <\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{6}-\\sqrt{2}}{4}=0.25881904510252\\)<br \/>\n\\(\\text{cos}(75\u00b0)=0.25881904510252\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt matches up, so we did something right! <\/p>\n<h3><span id=\"Tangent\" class=\"m-toc-anchor\"><\/span>Tangent<\/h3>\n<p>\nThere\u2019s also a sum identity for tangent:<\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(\\alpha + \\beta)=\\)<span style=\"font-size: 120%;\">\\(\\frac{\\text{tan}\\alpha + \\text{tan}\\beta}{1-\\text{tan}\\alpha\\text{ tan}\\beta}\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nLet\u2019s use this identity to answer this question: <\/p>\n<p>What is the exact value of the tangent of 75 degrees? <\/p>\n<p>In order to figure this out, we need to find the tangent of 30 degrees and the tangent of 45 degrees. But that information isn\u2019t on our unit circle. Fortunately, we can find it easily enough because we know that: <\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(\\alpha)=\\)<span style=\"font-size: 120%;\">\\(\\frac{\\text{sin }\\alpha}{\\text{cos }\\beta}\\)<\/span><\/div>\n<p>\n&nbsp;<br \/>\nWe could do that twice for each of the two angles we need and then plug it into the sum identity for tangent, but since we just found the exact value for the sine and cosine of 75 degrees by using the sum identities for both sine and cosine, we can plug those answers directly into this simpler identity to find our answer:<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(75\u00b0)=\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\)<br \/>\n\\(\\text{cos}(75\u00b0)=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo if we want to know the \\(\\text{tan}(75\u00b0)\\), we simply plug in the \\(\\text{sin}(75\u00b0)\\) and then divide it by the \\(\\text{cos}(75\u00b0)\\).<\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(75\u00b0)=\\frac{\\text{sin}(75\u00b0)}{\\text{cos}(75\u00b0)}=\\frac{\\frac{\\sqrt{2}+\\sqrt{6}}{4}}{\\frac{\\sqrt{6}-\\sqrt{2}}{4}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, to simplify this, we can rewrite this as division of fractions.<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\frac{\\sqrt{2}+\\sqrt{6}}{4}}{\\frac{\\sqrt{6}-\\sqrt{2}}{4}}=\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\div \\frac{\\sqrt{6}-\\sqrt{2}}{4}=\\frac{\\sqrt{2}+\\sqrt{6}}{4} \\frac{4}{\\sqrt{6}-\\sqrt{2}}=\\frac{\\sqrt{2}+\\sqrt{6}}{\\sqrt{6}-\\sqrt{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow remember, when we divide fractions, we flip the second fraction and multiply them.<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\div \\frac{\\sqrt{6}-\\sqrt{2}}{4}=\\frac{\\sqrt{2}+\\sqrt{6}}{4}\\cdot  \\frac{4}{\\sqrt{6}-\\sqrt{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can multiply these together.<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{2}+\\sqrt{6}}{4} \\cdot  \\frac{4}{\\sqrt{6}-\\sqrt{2}}=\\frac{\\sqrt{2}+\\sqrt{6}}{\\sqrt{6}-\\sqrt{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we didn\u2019t know the sine and cosine of 75 degrees, it would make sense to use our sum identity for tangent. So let\u2019s walk through what that would look like.<\/p>\n<p>As usual, we start by writing down all the information we\u2019re going to need. In this case, we have to use the fact that \\(tan=\\frac{sin}{cos}\\) to find our values for the tangent of 30 and 45 degrees. <\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(30\u00b0)=\\frac{\\text{sin}(30\u00b0)}{\\text{cos}(30\u00b0)}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis comes from rationalizing the denominator which is something we&#8217;ll cover a little bit more later on in this video.<\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(45\u00b0)=\\frac{\\text{sin}(45\u00b0)}{\\text{cos}(45\u00b0)}=\\frac{\\frac{\\sqrt{2}}{2}}{\\frac{\\sqrt{2}}{2}}=1\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, we write out our tangent sum identity.<\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(\\alpha+\\beta)=\\frac{\\text{tan}(\\alpha)+\\text{tan}(\\beta)}{1-\\text{tan}(\\alpha)\\text{tan}(\\beta)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we\u2019ll substitute 30 degrees for \\(\\alpha\\), 45 degrees for \\(\\beta\\), and then substitute all of our tangent values into our formula. <\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(30\u00b0+45\u00b0)=\\frac{\\text{tan}(30\u00b0)+\\text{tan}(45\u00b0)}{1-\\text{tan}(30\u00b0)\\text{tan}(45\u00b0)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, let\u2019s plug in our tangent values.<\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(30+45)=\\frac{\\frac{\\sqrt{3}}{3}+1}{1-(\\frac{\\sqrt{3}}{3})(1)}\\)\\(=\\frac{\\frac{\\sqrt{3}}{3}+\\frac{3}{3}}{1-\\frac{\\sqrt{3}}{3}}\\)\\(=\\frac{\\frac{\\sqrt{3}+3}{3}}{\\frac{3-\\sqrt{3}}{3}}= \\frac{\\sqrt{3}+3}{3}\\div \\frac{3-\\sqrt{3}}{3}\\)\\(=\\frac{\\sqrt{3}+3}{3}\\cdot \\frac{3}{3-\\sqrt{3}}\\)\\(=\\frac{\\sqrt{3}+3}{3-\\sqrt{3}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNotice that the two answers for our two different methods don\u2019t appear to be the same. But if we enter them into a calculator, we\u2019ll get the same value: <\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{3}+3}{3-\\sqrt{3}}=3.7320508075\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{2}+\\sqrt{6}}{\\sqrt{6}-\\sqrt{2}}=3.7320508075\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis is the same value we get when we ask the calculator for the tan of 75 degrees too! <\/p>\n<div class=\"examplesentence\">\\(\\text{tan}(75\u00b0)=3.7320508075\\)<\/div>\n<p>\n&nbsp;<\/p>\n<h3><span id=\"Simplifying_Answers\" class=\"m-toc-anchor\"><\/span>Simplifying Answers<\/h3>\n<p>\nThe reason the answers look different from each other is that they haven\u2019t been fully simplified. If you\u2019ll notice, there\u2019s a radical in the denominator. To fix both answers, we need to multiply 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. Then, we can simplify even further and see if we can make the answers look the same.<\/p>\n<p>Let\u2019s start with the first one. First, we need to make a conjugate version of the denominator and multiply the answer. The <strong>conjugate<\/strong> means that the sign in between the two terms is going to change. <\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{2}+\\sqrt{6}}{\\sqrt{6}-\\sqrt{2}}\\cdot \\frac{\\sqrt{6}+\\sqrt{2}}{\\sqrt{6}+\\sqrt{2}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis fraction simplifies to 1, so we&#8217;re not changing the value of our fraction at all. Now we can use <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/multiplying-terms-using-the-foil-method\/\">FOIL<\/a> to multiply our numerators and denominators together.<\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{12}+\\sqrt{4}+\\sqrt{36}+\\sqrt{12}}{\\sqrt{36}+\\sqrt{12}+(-\\sqrt{12})+(-\\sqrt{4})}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we&#8217;ll simplify our radicals a little bit.<\/p>\n<div class=\"examplesentence\">\\(\\frac{2\\sqrt{3}+2+6+2\\sqrt{3}}{6+2\\sqrt{3}-2\\sqrt{3}-2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can combine like terms in our numerator.<\/p>\n<div class=\"examplesentence\">\\(\\frac{4\\sqrt{3}+8}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom here we can factor 4 out of our numerator.<\/p>\n<div class=\"examplesentence\">\\(\\frac{4(\\sqrt{3}+2)}4\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur 4s will cancel each other out and we&#8217;ll be left with \\(\\sqrt{3}+2\\).<\/p>\n<p>Okay, now I\u2019ll do the same thing to the other one. Again, we&#8217;re going to multiply by fraction that has the conjugate in both the numerator and denominator. So the conjugate of \\(3-\\sqrt{3}\\) is \\(3+\\sqrt{3}\\). <\/p>\n<div class=\"examplesentence\">\\(\\frac{\\sqrt{3}+3}{3-\\sqrt{3}}\\cdot \\frac{3+\\sqrt{3}}{3+\\sqrt{3}}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we&#8217;re going to FOIL our numerators and denominators, and simplify our roots.<\/p>\n<div class=\"examplesentence\">\\(\\frac{3\\sqrt{3}+\\sqrt{9}+9+3\\sqrt{3}}{9+3\\sqrt{3}+(-3\\sqrt{3})+(-\\sqrt{9})}=\\frac{3\\sqrt{3}+3+9+3\\sqrt{3}}{9-3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can combine like terms in our numerator and denominator, and factor a 6 out of the numerator to simplify.<\/p>\n<div class=\"examplesentence\">\\(\\frac{6\\sqrt{3}+12}{6}=\\frac{6(\\sqrt{3}+2)}{6}=2+\\sqrt{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo we can see that both of our values are equal to the same thing, even though originally they looked a little bit different. And this is a much nicer number. So rationalizing the denominator can sometimes be really useful.<\/p>\n<h2><span id=\"Difference_Identities\" class=\"m-toc-anchor\"><\/span>Difference Identities<\/h2>\n<p>\nWe also have identities to find the <em>difference<\/em> of two angles rather than the sum. These are the difference identities for sine, cosine, and tangent:<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(\\alpha-\\beta)=\\text{sin}(\\alpha) \\text{cos}(\\beta)-\\text{cos}(\\alpha) \\text{sin}(\\beta)\\)<br \/>\n\\(\\text{cos}(\\alpha-\\beta)=\\text{cos}(\\alpha) \\text{cos}(\\beta)+\\text{sin}(\\alpha) \\text{sin}(\\beta)\\)<br \/>\n\\(\\text{tan}(\\alpha-\\beta)=\\frac{\\text{tan}(\\alpha) &#8211; \\text{tan}(\\beta)}{1 + \\text{tan} (\\alpha) \\text{tan}(\\beta)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe use these the same way we use the sum identities, but they enable us to find some smaller angles. For instance, we could use the sine difference identity to find the sine of 15 degrees, using 45 degrees as \\(\\alpha\\) and 30 degrees as \\(\\beta\\). <\/p>\n<p>Once again, we start by gathering the information we need and defining our \\(\\alpha\\) and \\(\\beta\\) angles to keep things organized and minimize the risk of a mistake. So for this problem, we&#8217;re saying that \\(\\alpha=45\u00b0\\). Since we&#8217;re looking for a smaller angle, our alpha has to be bigger than our beta. So \\(\\beta=30\u00b0\\).<\/p>\n<div class=\"examplesentence\">\\(\\alpha=45\u00b0~~~~~~~~\\beta=30\u00b0\\)<br \/>\n\\(\\text{sin}(45\u00b0)=\\frac{\\sqrt{2}}{2}~~~~~~~~\\text{sin}(30\u00b0)=\\frac{1}{2}\\)<br \/>\n\\(\\text{cos}(45\u00b0)=\\frac{\\sqrt{2}}{2}~~~~~~~~\\text{cos}(30\u00b0)=\\frac{\\sqrt{3}}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we&#8217;re going to write out our sum difference identity. <\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(\\alpha-\\beta)=\\text{sin}(\\alpha)\\text{cos}(\\beta)-\\text{cos}(\\alpha)\\text{sin}(\\beta)\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we simply substitute all of the appropriate values into the formula.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(45\u00b0-30\u00b0)=\\text{sin}(45\u00b0)\\text{cos}(30\u00b0)-\\text{cos}(45\u00b0)\\text{sin}(30\u00b0)\\)<br \/>\n\\(\\text{sin}(75\u00b0)=(\\frac{\\sqrt{2}}{2})(\\frac{\\sqrt{3}}{2})-(\\frac{\\sqrt{2}}{2})(\\frac{1}{2})\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can multiply and simplify a little bit.<\/p>\n<div class=\"examplesentence\">\\(\\text{sin}(15\u00b0)=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis time there is no radical in our denominator so we don\u2019t have to rationalize it.<\/p>\n<p>We can check this value on a calculator by finding \\(\\text{sin}(15\u00b0)\\). This gives us an approximate value of 0.259.<\/p>\n<p>Remember, our answer with radicals is the exact value for \\(\\text{sin}(15\u00b0)\\).<\/p>\n<p>The cosine and tangent difference identities work the same way. We simply plug in our values and enjoy working with radicals until we have our answer. <\/p>\n<hr>\n<p>\nI hope that this video helped you better understand the sum and difference identities of trig functions. Thanks for watching, and happy studying!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.onlinemathlearning.com\/sum-identities.html\"target=\"_blank\">\u201cSum and Difference Identities (Video Lessons, Examples and Solutions).\u201d n.d.<\/a><\/li>\n<li><a href=\"https:\/\/www.mathopenref.com\/trigidentities.html\"target=\"_blank\">\u201cTrigonometry Identities &#8211; Math Open Reference.\u201d n.d.<\/a><\/li>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/trigonometry\/\">Return to Trigonometry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Trigonometry Videos<\/p>\n","protected":false},"author":22,"featured_media":121195,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-121192","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-unit-circle-videos","7":"page_type-video","8":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/121192","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=121192"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/121192\/revisions"}],"predecessor-version":[{"id":261973,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/121192\/revisions\/261973"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/121195"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=121192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}