{"id":2006,"date":"2025-07-31T23:36:28","date_gmt":"2025-07-31T23:36:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2006"},"modified":"2025-10-14T21:19:33","modified_gmt":"2025-10-14T21:19:33","slug":"sum-to-product-and-product-to-sum-formulas-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sum-to-product-and-product-to-sum-formulas-learn-it-2\/","title":{"raw":"Sum-to-Product and Product-to-Sum Formulas: Learn It 2","rendered":"Sum-to-Product and Product-to-Sum Formulas: Learn It 2"},"content":{"raw":"

Expressing Sums as Products<\/h2>\r\nSome problems require the reverse of the process we just used. The sum-to-product formulas<\/strong> allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine<\/strong>. Let [latex]\\frac{u+v}{2}=\\alpha [\/latex] and [latex]\\frac{u-v}{2}=\\beta [\/latex].\r\n\r\nThen,\r\n
[latex]\\begin{align}\\alpha +\\beta &=\\frac{u+v}{2}+\\frac{u-v}{2} \\\\ &=\\frac{2u}{2} \\\\ &=u \\\\ \\text{ } \\\\ \\alpha -\\beta &=\\frac{u+v}{2}-\\frac{u-v}{2} \\\\ &=\\frac{2v}{2} \\\\ &=v \\end{align}[\/latex]<\/div>\r\nThus, replacing [latex]\\alpha [\/latex] and [latex]\\beta [\/latex] in the product-to-sum formula with the substitute expressions, we have\r\n
[latex]\\begin{align}&\\sin \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right] \\\\ &\\sin \\left(\\frac{u+v}{2}\\right)\\cos \\left(\\frac{u-v}{2}\\right)=\\frac{1}{2}\\left[\\sin u+\\sin v\\right]&& \\text{Substitute for}\\left(\\alpha +\\beta \\right)\\text{ and }\\left(\\alpha -\\beta \\right) \\\\ &2\\sin \\left(\\frac{u+v}{2}\\right)\\cos \\left(\\frac{u-v}{2}\\right)=\\sin u+\\sin v \\end{align}[\/latex]<\/div>\r\nThe other sum-to-product identities are derived similarly.\r\n\r\n
\r\n

sum-to-product formulas<\/h3>\r\nThe sum-to-product formulas<\/strong> are as follows:\r\n

[latex]\\sin \\alpha +\\sin \\beta =2\\sin \\left(\\frac{\\alpha +\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\r\n

[latex]\\sin \\alpha -\\sin \\beta =2\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha +\\beta }{2}\\right)[\/latex]<\/p>\r\n

[latex]\\cos \\alpha -\\cos \\beta =-2\\sin \\left(\\frac{\\alpha +\\beta }{2}\\right)\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\r\n

[latex]\\cos \\alpha +\\cos \\beta =2\\cos \\left(\\frac{\\alpha +\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\r\n\r\n<\/section>

Write the following difference of sines expression as a product: [latex]\\sin \\left(4\\theta \\right)-\\sin \\left(2\\theta \\right)[\/latex].[reveal-answer q=\"627723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"627723\"]We begin by writing the formula for the difference of sines.\r\n

[latex]\\sin \\alpha -\\sin \\beta =2\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha +\\beta }{2}\\right)[\/latex]<\/p>\r\nSubstitute the values into the formula, and simplify.\r\n

[latex]\\begin{align}\\sin \\left(4\\theta \\right)-\\sin \\left(2\\theta \\right)&=2\\sin \\left(\\frac{4\\theta -2\\theta }{2}\\right)\\cos \\left(\\frac{4\\theta +2\\theta }{2}\\right) \\\\ &=2\\sin \\left(\\frac{2\\theta }{2}\\right)\\cos \\left(\\frac{6\\theta }{2}\\right) \\\\ &=2\\sin \\theta \\cos \\left(3\\theta \\right) \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n
\r\n\r\nUse the sum-to-product formula to write the sum as a product: [latex]\\sin \\left(3\\theta \\right)+\\sin \\left(\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"34084\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"34084\"]\r\n\r\n[latex]2\\sin \\left(2\\theta \\right)\\cos \\left(\\theta \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
[ohm_question hide_question_numbers=1]4632[\/ohm_question]<\/section>
Evaluate [latex]\\cos \\left({15}^{\\circ }\\right)-\\cos \\left({75}^{\\circ }\\right)[\/latex].[reveal-answer q=\"296492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296492\"]We begin by writing the formula for the difference of cosines.\r\n

[latex]\\cos \\alpha -\\cos \\beta =-2\\sin \\left(\\frac{\\alpha +\\beta }{2}\\right)\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\r\nThen we substitute the given angles and simplify.\r\n

[latex]\\begin{align}\\cos \\left({15}^{\\circ }\\right)-\\cos \\left({75}^{\\circ }\\right)&=-2\\sin \\left(\\frac{{15}^{\\circ }+{75}^{\\circ }}{2}\\right)\\sin \\left(\\frac{{15}^{\\circ }-{75}^{\\circ }}{2}\\right) \\\\ &=-2\\sin \\left({45}^{\\circ }\\right)\\sin \\left(-{30}^{\\circ }\\right) \\\\ &=-2\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(-\\frac{1}{2}\\right) \\\\ &=\\frac{\\sqrt{2}}{2}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Prove the identity:\r\n

[latex]\\frac{\\cos \\left(4t\\right)-\\cos \\left(2t\\right)}{\\sin \\left(4t\\right)+\\sin \\left(2t\\right)}=-\\tan t[\/latex]<\/p>\r\n[reveal-answer q=\"514870\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"514870\"]\r\n\r\nWe will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.\r\n

[latex]\\begin{align}\\frac{\\cos \\left(4t\\right)-\\cos \\left(2t\\right)}{\\sin \\left(4t\\right)+\\sin \\left(2t\\right)}&=\\frac{-2\\sin \\left(\\frac{4t+2t}{2}\\right)\\sin \\left(\\frac{4t - 2t}{2}\\right)}{2\\sin \\left(\\frac{4t+2t}{2}\\right)\\cos \\left(\\frac{4t - 2t}{2}\\right)} \\\\ &=\\frac{-2\\sin \\left(3t\\right)\\sin t}{2\\sin \\left(3t\\right)\\cos t} \\\\ &=\\frac{-2\\sin \\left(3t\\right)\\sin t}{2\\sin \\left(3t\\right)\\cos t} \\\\ &=-\\frac{\\sin t}{\\cos t} \\\\ &=-\\tan t \\end{align}[\/latex]<\/p>\r\n\r\n

Analysis of the Solution<\/h4>\r\nRecall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Expressing Sums as Products<\/h2>\n

Some problems require the reverse of the process we just used. The sum-to-product formulas<\/strong> allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine<\/strong>. Let [latex]\\frac{u+v}{2}=\\alpha[\/latex] and [latex]\\frac{u-v}{2}=\\beta[\/latex].<\/p>\n

Then,<\/p>\n

[latex]\\begin{align}\\alpha +\\beta &=\\frac{u+v}{2}+\\frac{u-v}{2} \\\\ &=\\frac{2u}{2} \\\\ &=u \\\\ \\text{ } \\\\ \\alpha -\\beta &=\\frac{u+v}{2}-\\frac{u-v}{2} \\\\ &=\\frac{2v}{2} \\\\ &=v \\end{align}[\/latex]<\/div>\n

Thus, replacing [latex]\\alpha[\/latex] and [latex]\\beta[\/latex] in the product-to-sum formula with the substitute expressions, we have<\/p>\n

[latex]\\begin{align}&\\sin \\alpha \\cos \\beta =\\frac{1}{2}\\left[\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)\\right] \\\\ &\\sin \\left(\\frac{u+v}{2}\\right)\\cos \\left(\\frac{u-v}{2}\\right)=\\frac{1}{2}\\left[\\sin u+\\sin v\\right]&& \\text{Substitute for}\\left(\\alpha +\\beta \\right)\\text{ and }\\left(\\alpha -\\beta \\right) \\\\ &2\\sin \\left(\\frac{u+v}{2}\\right)\\cos \\left(\\frac{u-v}{2}\\right)=\\sin u+\\sin v \\end{align}[\/latex]<\/div>\n

The other sum-to-product identities are derived similarly.<\/p>\n

\n

sum-to-product formulas<\/h3>\n

The sum-to-product formulas<\/strong> are as follows:<\/p>\n

[latex]\\sin \\alpha +\\sin \\beta =2\\sin \\left(\\frac{\\alpha +\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\n

[latex]\\sin \\alpha -\\sin \\beta =2\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha +\\beta }{2}\\right)[\/latex]<\/p>\n

[latex]\\cos \\alpha -\\cos \\beta =-2\\sin \\left(\\frac{\\alpha +\\beta }{2}\\right)\\sin \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\n

[latex]\\cos \\alpha +\\cos \\beta =2\\cos \\left(\\frac{\\alpha +\\beta }{2}\\right)\\cos \\left(\\frac{\\alpha -\\beta }{2}\\right)[\/latex]<\/p>\n<\/section>\n

Write the following difference of sines expression as a product: [latex]\\sin \\left(4\\theta \\right)-\\sin \\left(2\\theta \\right)[\/latex].<\/p>\n