{"id":203,"date":"2025-02-13T22:44:59","date_gmt":"2025-02-13T22:44:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sum-and-difference-identities\/"},"modified":"2025-10-14T18:21:41","modified_gmt":"2025-10-14T18:21:41","slug":"sum-and-difference-identities","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sum-and-difference-identities\/","title":{"raw":"Sum and Difference Identities: Learn It 1","rendered":"Sum and Difference Identities: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use sum and difference formulas for sine, cosine, and tangent<\/li>\r\n \t<li style=\"font-weight: 400;\">Use sum and difference formulas to verify identities.<\/li>\r\n<\/ul>\r\n<\/section>Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">The unit circle contains all of the special angles along with their cosine and sine values. Remember that each ordered pair is in the form [latex](\\cos x,\\sin x)[\/latex].<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164036\/CNX_Precalc_Figure_07_01_0042.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B. \" width=\"975\" height=\"638\" \/><\/section>\r\n<h2>Use sum and difference formulas for cosine<\/h2>\r\nWe will begin with the <strong>sum and difference formulas for cosine<\/strong>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>sum and difference of cosine<\/h3>\r\n<strong>Sum formula for cosine<\/strong>\r\n\r\n[latex]\\cos \\left(\\alpha +\\beta \\right) = \\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex]\r\n\r\n&nbsp;\r\n\r\n<strong>Difference formula for cosine<\/strong>\r\n\r\n[latex]\\cos \\left(\\alpha -\\beta \\right) = \\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta [\/latex]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\"><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">First, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. Point [latex]P[\/latex] is at an angle [latex]\\alpha [\/latex] from the positive <\/span><em style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">x-<\/em><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">axis with coordinates [latex]\\left(\\cos \\alpha ,\\sin \\alpha \\right)[\/latex] and point [latex]Q[\/latex] is at an angle of [latex]\\beta [\/latex] from the positive <\/span><em style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">x-<\/em><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">axis with coordinates [latex]\\left(\\cos \\beta ,\\sin \\beta \\right)[\/latex]. Note the measure of angle [latex]POQ[\/latex] is [latex]\\alpha -\\beta [\/latex].<\/span>Label two more points: [latex]A[\/latex] at an angle of [latex]\\left(\\alpha -\\beta \\right)[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\left(\\alpha -\\beta \\right),\\sin \\left(\\alpha -\\beta \\right)\\right)[\/latex]; and point [latex]B[\/latex] with coordinates [latex]\\left(1,0\\right)[\/latex]. Triangle [latex]POQ[\/latex] is a rotation of triangle [latex]AOB[\/latex] and thus the distance from [latex]P[\/latex] to [latex]Q[\/latex] is the same as the distance from [latex]A[\/latex] to [latex]B[\/latex].\r\n<figure id=\"Figure_07_02_002\" class=\"small\"><span id=\"fs-id1130636\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164038\/CNX_Precalc_Figure_07_02_0022.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" \/><\/span><\/figure>\r\n<p style=\"text-align: center;\">We can find the distance from [latex]P[\/latex] to [latex]Q[\/latex] using the <strong>distance formula<\/strong>.<\/p>\r\n\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{d}_{PQ}&amp;=\\sqrt{{\\left(\\cos \\alpha -\\cos \\beta \\right)}^{2}+{\\left(\\sin \\alpha -\\sin \\beta \\right)}^{2}} \\\\ &amp;=\\sqrt{{\\cos }^{2}\\alpha -2\\cos \\alpha \\cos \\beta +{\\cos }^{2}\\beta +{\\sin }^{2}\\alpha -2\\sin \\alpha \\sin \\beta +{\\sin }^{2}\\beta } \\end{align}[\/latex]<\/div>\r\nThen we apply the <strong>Pythagorean identity<\/strong> and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} &amp;=\\sqrt{\\left({\\cos }^{2}\\alpha +{\\sin }^{2}\\alpha \\right)+\\left({\\cos }^{2}\\beta +{\\sin }^{2}\\beta \\right)-2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &amp;=\\sqrt{1+1 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &amp;=\\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\end{align}[\/latex]<\/div>\r\nSimilarly, using the distance formula we can find the distance from [latex]A[\/latex] to [latex]B[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{d}_{AB}&amp;=\\sqrt{{\\left(\\cos \\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\sin \\left(\\alpha -\\beta \\right)-0\\right)}^{2}} \\\\ &amp;=\\sqrt{{\\cos }^{2}\\left(\\alpha -\\beta \\right)-2\\cos \\left(\\alpha -\\beta \\right)+1+{\\sin }^{2}\\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\r\nApplying the Pythagorean identity and simplifying we get:\r\n<div style=\"text-align: center;\">[latex]\\begin{align} &amp;=\\sqrt{\\left({\\cos }^{2}\\left(\\alpha -\\beta \\right)+{\\sin }^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &amp;=\\sqrt{1 - 2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &amp;=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\r\nBecause the two distances are the same, we set them equal to each other and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} \\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta }&amp;=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\\\ 2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta &amp;=2 - 2\\cos \\left(\\alpha -\\beta \\right) \\end{align}[\/latex]<\/div>\r\nFinally, we subtract [latex]2[\/latex] from both sides and divide both sides by [latex]-2[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =\\cos \\left(\\alpha -\\beta \\right)\\end{align}[\/latex]<\/div>\r\nThus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given two angles, find the cosine of the difference between the angles.\r\n<\/strong>\r\n<ol id=\"fs-id1374215\">\r\n \t<li>Write the difference formula for cosine.<\/li>\r\n \t<li>Substitute the values of the given angles into the formula.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Using the formula for the cosine of the difference of two angles, find the exact value of [latex]\\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)[\/latex].[reveal-answer q=\"956915\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"956915\"]Use the formula for the cosine of the difference of two angles. We have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)&amp;=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\\\ \\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)&amp;=\\cos \\left(\\frac{5\\pi }{4}\\right)\\cos \\left(\\frac{\\pi }{6}\\right)+\\sin \\left(\\frac{5\\pi }{4}\\right)\\sin \\left(\\frac{\\pi }{6}\\right) \\\\ &amp;=\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right) \\\\ &amp;=-\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &amp;=\\frac{-\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nFind the exact value of [latex]\\cos \\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"700317\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"700317\"]\r\n\r\n[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the exact value of [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex].[reveal-answer q=\"617422\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"617422\"]As [latex]{75}^{\\circ }={45}^{\\circ }+{30}^{\\circ }[\/latex], we can evaluate [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex] as [latex]\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)&amp;=\\cos \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\sin \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\\\ &amp;=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &amp;=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &amp;=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">Find the exact value of [latex]\\cos \\left({105}^{\\circ }\\right)[\/latex].[reveal-answer q=\"683323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"683323\"][latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]173445[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li style=\"font-weight: 400;\">Use sum and difference formulas for sine, cosine, and tangent<\/li>\n<li style=\"font-weight: 400;\">Use sum and difference formulas to verify identities.<\/li>\n<\/ul>\n<\/section>\n<p>Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">The unit circle contains all of the special angles along with their cosine and sine values. Remember that each ordered pair is in the form [latex](\\cos x,\\sin x)[\/latex].<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164036\/CNX_Precalc_Figure_07_01_0042.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"975\" height=\"638\" \/><\/section>\n<h2>Use sum and difference formulas for cosine<\/h2>\n<p>We will begin with the <strong>sum and difference formulas for cosine<\/strong>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>sum and difference of cosine<\/h3>\n<p><strong>Sum formula for cosine<\/strong><\/p>\n<p>[latex]\\cos \\left(\\alpha +\\beta \\right) = \\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Difference formula for cosine<\/strong><\/p>\n<p>[latex]\\cos \\left(\\alpha -\\beta \\right) = \\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\"><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">First, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. Point [latex]P[\/latex] is at an angle [latex]\\alpha[\/latex] from the positive <\/span><em style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">x-<\/em><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">axis with coordinates [latex]\\left(\\cos \\alpha ,\\sin \\alpha \\right)[\/latex] and point [latex]Q[\/latex] is at an angle of [latex]\\beta[\/latex] from the positive <\/span><em style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">x-<\/em><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">axis with coordinates [latex]\\left(\\cos \\beta ,\\sin \\beta \\right)[\/latex]. Note the measure of angle [latex]POQ[\/latex] is [latex]\\alpha -\\beta[\/latex].<\/span>Label two more points: [latex]A[\/latex] at an angle of [latex]\\left(\\alpha -\\beta \\right)[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\left(\\alpha -\\beta \\right),\\sin \\left(\\alpha -\\beta \\right)\\right)[\/latex]; and point [latex]B[\/latex] with coordinates [latex]\\left(1,0\\right)[\/latex]. Triangle [latex]POQ[\/latex] is a rotation of triangle [latex]AOB[\/latex] and thus the distance from [latex]P[\/latex] to [latex]Q[\/latex] is the same as the distance from [latex]A[\/latex] to [latex]B[\/latex].<\/p>\n<figure id=\"Figure_07_02_002\" class=\"small\"><span id=\"fs-id1130636\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164038\/CNX_Precalc_Figure_07_02_0022.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" \/><\/span><\/figure>\n<p style=\"text-align: center;\">We can find the distance from [latex]P[\/latex] to [latex]Q[\/latex] using the <strong>distance formula<\/strong>.<\/p>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{align}{d}_{PQ}&=\\sqrt{{\\left(\\cos \\alpha -\\cos \\beta \\right)}^{2}+{\\left(\\sin \\alpha -\\sin \\beta \\right)}^{2}} \\\\ &=\\sqrt{{\\cos }^{2}\\alpha -2\\cos \\alpha \\cos \\beta +{\\cos }^{2}\\beta +{\\sin }^{2}\\alpha -2\\sin \\alpha \\sin \\beta +{\\sin }^{2}\\beta } \\end{align}[\/latex]<\/div>\n<p>Then we apply the <strong>Pythagorean identity<\/strong> and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align} &=\\sqrt{\\left({\\cos }^{2}\\alpha +{\\sin }^{2}\\alpha \\right)+\\left({\\cos }^{2}\\beta +{\\sin }^{2}\\beta \\right)-2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &=\\sqrt{1+1 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &=\\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\end{align}[\/latex]<\/div>\n<p>Similarly, using the distance formula we can find the distance from [latex]A[\/latex] to [latex]B[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{d}_{AB}&=\\sqrt{{\\left(\\cos \\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\sin \\left(\\alpha -\\beta \\right)-0\\right)}^{2}} \\\\ &=\\sqrt{{\\cos }^{2}\\left(\\alpha -\\beta \\right)-2\\cos \\left(\\alpha -\\beta \\right)+1+{\\sin }^{2}\\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\n<p>Applying the Pythagorean identity and simplifying we get:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align} &=\\sqrt{\\left({\\cos }^{2}\\left(\\alpha -\\beta \\right)+{\\sin }^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &=\\sqrt{1 - 2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\n<p>Because the two distances are the same, we set them equal to each other and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align} \\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta }&=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\\\ 2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta &=2 - 2\\cos \\left(\\alpha -\\beta \\right) \\end{align}[\/latex]<\/div>\n<p>Finally, we subtract [latex]2[\/latex] from both sides and divide both sides by [latex]-2[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =\\cos \\left(\\alpha -\\beta \\right)\\end{align}[\/latex]<\/div>\n<p>Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given two angles, find the cosine of the difference between the angles.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1374215\">\n<li>Write the difference formula for cosine.<\/li>\n<li>Substitute the values of the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Using the formula for the cosine of the difference of two angles, find the exact value of [latex]\\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q956915\">Show Solution<\/button><\/p>\n<div id=\"q956915\" class=\"hidden-answer\" style=\"display: none\">Use the formula for the cosine of the difference of two angles. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)&=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\\\ \\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)&=\\cos \\left(\\frac{5\\pi }{4}\\right)\\cos \\left(\\frac{\\pi }{6}\\right)+\\sin \\left(\\frac{5\\pi }{4}\\right)\\sin \\left(\\frac{\\pi }{6}\\right) \\\\ &=\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right) \\\\ &=-\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &=\\frac{-\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>Find the exact value of [latex]\\cos \\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q700317\">Show Solution<\/button><\/p>\n<div id=\"q700317\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the exact value of [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q617422\">Show Solution<\/button><\/p>\n<div id=\"q617422\" class=\"hidden-answer\" style=\"display: none\">As [latex]{75}^{\\circ }={45}^{\\circ }+{30}^{\\circ }[\/latex], we can evaluate [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex] as [latex]\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)&=\\cos \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\sin \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\\\ &=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">Find the exact value of [latex]\\cos \\left({105}^{\\circ }\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q683323\">Show Solution<\/button><\/p>\n<div id=\"q683323\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm173445\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173445&theme=lumen&iframe_resize_id=ohm173445&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":6,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":201,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/203"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/203\/revisions"}],"predecessor-version":[{"id":4667,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/203\/revisions\/4667"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/201"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/203\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=203"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=203"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=203"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}