{"id":1993,"date":"2025-07-31T23:26:45","date_gmt":"2025-07-31T23:26:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1993"},"modified":"2025-10-14T18:40:54","modified_gmt":"2025-10-14T18:40:54","slug":"double-angle-half-angle-and-reduction-formulas-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas-learn-it-3\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas: Learn It 3","rendered":"Double Angle, Half Angle, and Reduction Formulas: Learn It 3"},"content":{"raw":"

Use Reduction Formulas to Simplify an Expression<\/h2>\r\nThe double-angle formulas can be used to derive the reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.\r\n\r\nWe can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta [\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]\r\n
[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\r\nNext, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]\r\n
[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\r\nThe last reduction formula is derived by writing tangent in terms of sine and cosine:\r\n
[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&& \\text{Substitute the reduction formulas.} \\\\ &=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\r\n
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reduction formulas<\/h3>\r\n[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]\r\n\r\n<\/section><\/div>\r\n
Write an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.[reveal-answer q=\"109691\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109691\"]We will apply the reduction formula for cosine twice.\r\n

[latex]\\begin{align}{\\cos }^{4}x&={\\left({\\cos }^{2}x\\right)}^{2} \\\\ &={\\left(\\frac{1+\\cos \\left(2x\\right)}{2}\\right)}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}\\left(1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right) \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1+\\cos \\left(2\\left(2x\\right)\\right)}{2}\\right) && {\\text{ Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}+\\frac{1}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{3}{8}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nThe solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Use the power-reducing formulas to prove\r\n

[latex]{\\sin }^{3}\\left(2x\\right)=\\left[\\frac{1}{2}\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right][\/latex]<\/p>\r\n[reveal-answer q=\"828553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"828553\"]\r\n\r\nWe will work on simplifying the left side of the equation:\r\n

[latex]\\begin{align}{\\sin }^{3}\\left(2x\\right)&=\\left[\\sin \\left(2x\\right)\\right]\\left[{\\sin }^{2}\\left(2x\\right)\\right] \\\\ &=\\sin \\left(2x\\right)\\left[\\frac{1-\\cos \\left(4x\\right)}{2}\\right]&& \\text{Substitute the power-reduction formula}. \\\\ &=\\sin \\left(2x\\right)\\left(\\frac{1}{2}\\right)\\left[1-\\cos \\left(4x\\right)\\right] \\\\ &=\\frac{1}{2}\\left[\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right] \\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nNote that in this example, we substituted\r\n

[latex]\\frac{1-\\cos \\left(4x\\right)}{2}[\/latex]<\/div>\r\nfor [latex]{\\sin }^{2}\\left(2x\\right)[\/latex]. The formula states\r\n
[latex]{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2}[\/latex]<\/div>\r\nWe let [latex]\\theta =2x[\/latex], so [latex]2\\theta =4x[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
\r\n
\r\n\r\nUse the power-reducing formulas to prove that [latex]10{\\cos }^{4}x=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"387102\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387102\"]\r\n

[latex]\\begin{align}10{\\cos }^{4}x&=10{\\left({\\cos }^{2}x\\right)}^{2} \\\\ &=10{\\left[\\frac{1+\\cos \\left(2x\\right)}{2}\\right]}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}\\left[1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right] \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{4}\\left(\\frac{1+\\cos\\left( 2\\left(2x\\right)\\right)}{2}\\right)&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{8}+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{30}{8}+5\\cos \\left(2x\\right)+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>

[ohm_question hide_question_numbers=1]4533[\/ohm_question]<\/section>","rendered":"

Use Reduction Formulas to Simplify an Expression<\/h2>\n

The double-angle formulas can be used to derive the reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.<\/p>\n

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta[\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]<\/p>\n

[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\n

Next, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]<\/p>\n

[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\n

The last reduction formula is derived by writing tangent in terms of sine and cosine:<\/p>\n

[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&& \\text{Substitute the reduction formulas.} \\\\ &=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\n
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reduction formulas<\/h3>\n

[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/p>\n<\/section>\n<\/div>\n

Write an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.<\/p>\n