{"id":1987,"date":"2025-07-31T23:16:51","date_gmt":"2025-07-31T23:16:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1987"},"modified":"2025-10-14T18:31:43","modified_gmt":"2025-10-14T18:31:43","slug":"sum-and-difference-identities-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sum-and-difference-identities-learn-it-3\/","title":{"raw":"Sum and Difference Identities: Learn It 3","rendered":"Sum and Difference Identities: Learn It 3"},"content":{"raw":"

Use sum and difference formulas for tangent<\/h2>\r\nFinding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.\r\n\r\nFinding the sum of two angles formula for tangent involves taking the quotient of the sum formulas for sine and cosine and simplifying. Recall, [latex]\\tan x=\\frac{\\sin x}{\\cos x},\\cos x\\ne 0[\/latex].\r\n\r\nLet\u2019s derive the sum formula for tangent.\r\n
[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)} \\\\[1mm] &=\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta } \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Divide the numerator and denominator by cos}\\alpha \\text{cos}\\beta \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta}+\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta }-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Split the fractions.} \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha }{\\cos \\alpha }+\\frac{\\sin \\beta }{\\cos \\beta }}{1-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Cancel.} \\\\[1mm] &=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\end{align}[\/latex]<\/div>\r\nWe can derive the difference formula for tangent in a similar way.\r\n\r\n
\r\n

sum and difference formula for tangent<\/h3>\r\n

[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\r\n

[latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\r\n\r\n<\/section>

How To: Given two angles, find the tangent of the sum of the angles.\r\n<\/strong>\r\n
    \r\n \t
  1. Write the sum formula for tangent.<\/li>\r\n \t
  2. Substitute the given angles into the formula.<\/li>\r\n \t
  3. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>
    Find the exact value of [latex]\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)[\/latex].[reveal-answer q=\"796417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796417\"]Let\u2019s first write the sum formula for tangent and substitute the given angles into the formula.\r\n

    [latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&=\\frac{\\tan \\left(\\frac{\\pi }{6}\\right)+\\tan \\left(\\frac{\\pi }{4}\\right)}{1-\\left(\\tan \\left(\\frac{\\pi }{6}\\right)\\right)\\left(\\tan \\left(\\frac{\\pi }{4}\\right)\\right)} \\end{align}[\/latex]<\/p>\r\nNext, we determine the individual tangents within the formula:\r\n

    [latex]\\tan \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{\\sqrt{3}},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/p>\r\nSo we have\r\n

    [latex]\\begin{align}\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&=\\frac{\\frac{1}{\\sqrt{3}}+1}{1-\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(1\\right)} \\\\ &=\\frac{\\frac{1+\\sqrt{3}}{\\sqrt{3}}}{\\frac{\\sqrt{3}-1}{\\sqrt{3}}} \\\\ &=\\frac{1+\\sqrt{3}}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{\\sqrt{3}-1}\\right) \\\\ &=\\frac{\\sqrt{3}+1}{\\sqrt{3}-1} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Find the exact value of [latex]\\tan \\left(\\frac{2\\pi }{3}+\\frac{\\pi }{4}\\right)[\/latex].[reveal-answer q=\"962695\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962695\"][latex]\\frac{1-\\sqrt{3}}{1+\\sqrt{3}}[\/latex][\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]173454[\/ohm_question]<\/section>
    Given [latex]\\text{ }\\sin \\alpha =\\frac{3}{5},0<\\alpha <\\frac{\\pi }{2},\\cos \\beta =-\\frac{5}{13},\\pi <\\beta <\\frac{3\\pi }{2}[\/latex], find\r\n
      \r\n \t
    1. [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t
    2. [latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t
    3. [latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t
    4. [latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"622511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"622511\"]\r\n\r\nWe can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.\r\n
        \r\n \t
      1. To find [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex], we begin with [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]0<\\alpha <\\frac{\\pi }{2}[\/latex]. The side opposite [latex]\\alpha [\/latex] has length 3, the hypotenuse has length 5, and [latex]\\alpha [\/latex] is in the first quadrant. Using the Pythagorean Theorem, we can find the length of side [latex]a:[\/latex]\r\n
        \r\n\r\n[latex]\\begin{gathered}{a}^{2}+{3}^{2}={5}^{2} \\\\ {a}^{2}=16 \\\\ a=4 \\end{gathered}[\/latex]\r\n\r\n\"Diagram\r\n\r\n<\/div>\r\nSince [latex]\\cos \\beta =-\\frac{5}{13}[\/latex] and [latex]\\pi <\\beta <\\frac{3\\pi }{2}[\/latex], the side adjacent to [latex]\\beta [\/latex] is [latex]-5[\/latex], the hypotenuse is 13, and [latex]\\beta [\/latex] is in the third quadrant. Again, using the Pythagorean Theorem, we have<\/li>\r\n \t
      2. \r\n

        [latex]\\begin{align}{\\left(-5\\right)}^{2}+{a}^{2}&={13}^{2} \\\\ 25+{a}^{2}&=169 \\\\ {a}^{2}&=144\\\\ a&=\\pm 12 \\end{align}[\/latex]<\/p>\r\nSince [latex]\\beta [\/latex] is in the third quadrant, [latex]a=-12[\/latex].\r\n\r\n\"Diagram\r\n\r\nThe next step is finding the cosine of [latex]\\alpha [\/latex] and the sine of [latex]\\beta [\/latex]. The cosine of [latex]\\alpha [\/latex] is the adjacent side over the hypotenuse. We can find it from the triangle: [latex]\\cos \\alpha =\\frac{4}{5}[\/latex]. We can also find the sine of [latex]\\beta [\/latex] from the triangle, as opposite side over the hypotenuse: [latex]\\sin \\beta =-\\frac{12}{13}[\/latex]. Now we are ready to evaluate [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex].\r\n

        [latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)&=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\\\ &=\\left(\\frac{3}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &=-\\frac{15}{65}-\\frac{48}{65} \\\\ &=-\\frac{63}{65} \\end{align}[\/latex]<\/div><\/li>\r\n \t
      3. We can find [latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex] in a similar manner. We substitute the values according to the formula.\r\n
        [latex]\\begin{align}\\cos \\left(\\alpha +\\beta \\right)&=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\\\ &=\\left(\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)-\\left(\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &=-\\frac{20}{65}+\\frac{36}{65} \\\\ &=\\frac{16}{65} \\end{align}[\/latex]<\/div><\/li>\r\n \t
      4. For [latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex], if [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]\\cos \\alpha =\\frac{4}{5}[\/latex], then\r\n
        [latex]\\tan \\alpha =\\frac{\\frac{3}{5}}{\\frac{4}{5}}=\\frac{3}{4}[\/latex]<\/div>\r\nIf [latex]\\sin \\beta =-\\frac{12}{13}[\/latex] and [latex]\\cos \\beta =-\\frac{5}{13}[\/latex],\r\nthen\r\n
        [latex]\\tan \\beta =\\frac{\\frac{-12}{13}}{\\frac{-5}{13}}=\\frac{12}{5}[\/latex]<\/div>\r\nThen,\r\n
        [latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\\\ &=\\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &=\\frac{\\text{ }\\frac{63}{20}}{-\\frac{16}{20}} \\\\ &=-\\frac{63}{16} \\end{align}[\/latex]<\/div><\/li>\r\n \t
      5. To find [latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex], we have the values we need. We can substitute them in and evaluate.\r\n
        [latex]\\begin{align}\\tan \\left(\\alpha -\\beta \\right)&=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta } \\\\ &=\\frac{\\frac{3}{4}-\\frac{12}{5}}{1+\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &=\\frac{-\\frac{33}{20}}{\\frac{56}{20}} \\\\ &=-\\frac{33}{56}\\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n

        Analysis of the Solution<\/h4>\r\nA common mistake when addressing problems such as this one is that we may be tempted to think that [latex]\\alpha [\/latex] and [latex]\\beta [\/latex] are angles in the same triangle, which of course, they are not. Also note that\r\n
        [latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

        Use sum and difference formulas for tangent<\/h2>\n

        Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.<\/p>\n

        Finding the sum of two angles formula for tangent involves taking the quotient of the sum formulas for sine and cosine and simplifying. Recall, [latex]\\tan x=\\frac{\\sin x}{\\cos x},\\cos x\\ne 0[\/latex].<\/p>\n

        Let\u2019s derive the sum formula for tangent.<\/p>\n

        [latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)} \\\\[1mm] &=\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta } \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Divide the numerator and denominator by cos}\\alpha \\text{cos}\\beta \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta}+\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta }-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Split the fractions.} \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha }{\\cos \\alpha }+\\frac{\\sin \\beta }{\\cos \\beta }}{1-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Cancel.} \\\\[1mm] &=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\end{align}[\/latex]<\/div>\n

        We can derive the difference formula for tangent in a similar way.<\/p>\n

        \n

        sum and difference formula for tangent<\/h3>\n

        [latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\n

        [latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\n<\/section>\n

        How To: Given two angles, find the tangent of the sum of the angles.
        \n<\/strong><\/p>\n
          \n
        1. Write the sum formula for tangent.<\/li>\n
        2. Substitute the given angles into the formula.<\/li>\n
        3. Simplify.<\/li>\n<\/ol>\n<\/section>\n
          Find the exact value of [latex]\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n