{"id":1991,"date":"2025-07-31T23:25:27","date_gmt":"2025-07-31T23:25:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1991"},"modified":"2025-10-14T18:39:04","modified_gmt":"2025-10-14T18:39:04","slug":"double-angle-half-angle-and-reduction-formulas-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas-learn-it-1\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas: Learn It 2","rendered":"Double Angle, Half Angle, and Reduction Formulas: Learn It 2"},"content":{"raw":"

Using Double-Angle Formulas to Verify Identities<\/h2>\r\nEstablishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.\r\n\r\n
Establish the following identity using double-angle formulas:\r\n

[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\r\n[reveal-answer q=\"600157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"600157\"]\r\n\r\nWe will work on the right side of the equal sign and rewrite the expression until it matches the left side.\r\n

[latex]\\begin{align}{\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}&={\\sin }^{2}\\theta +2\\sin \\theta \\cos \\theta +{\\cos }^{2}\\theta \\\\ &=\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+2\\sin \\theta \\cos \\theta \\\\ &=1+2\\sin \\theta \\cos \\theta \\\\ &=1+\\sin \\left(2\\theta \\right)\\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nThis process is not complicated, as long as we recall the perfect square formula from algebra:\r\n

[latex]{\\left(a\\pm b\\right)}^{2}={a}^{2}\\pm 2ab+{b}^{2}[\/latex]<\/p>\r\nwhere [latex]a=\\sin \\theta [\/latex] and [latex]b=\\cos \\theta [\/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Establish the identity: [latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\cos \\left(2\\theta \\right)[\/latex].[reveal-answer q=\"178499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178499\"][latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)=\\cos \\left(2\\theta \\right)[\/latex][\/hidden-answer]<\/section>
Verify the identity:\r\n

[latex]\\tan \\left(2\\theta \\right)=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\n[reveal-answer q=\"395376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"395376\"]\r\n\r\nIn this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.\r\n

[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }&& \\text{Double-angle formula} \\\\ &=\\frac{2\\tan \\theta \\left(\\frac{1}{\\tan \\theta }\\right)}{\\left(1-{\\tan }^{2}\\theta \\right)\\left(\\frac{1}{\\tan \\theta }\\right)}&& \\text{Multiply by a term that results in desired numerator}. \\\\ &=\\frac{2}{\\frac{1}{\\tan \\theta }-\\frac{{\\tan }^{2}\\theta }{\\tan \\theta }} \\\\ &=\\frac{2}{\\cot \\theta -\\tan \\theta }&& \\text{Use reciprocal identity for }\\frac{1}{\\tan \\theta }.\\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nHere is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show\r\n

[latex]\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\nLet\u2019s work on the right side.\r\n

[latex]\\begin{align}\\frac{2}{\\cot \\theta -\\tan \\theta }&=\\frac{2}{\\frac{1}{\\tan \\theta }-\\tan \\theta }\\left(\\frac{\\tan \\theta }{\\tan \\theta }\\right) \\\\ &=\\frac{2\\tan \\theta }{\\frac{1}{\\cancel{\\tan \\theta }}\\left(\\cancel{\\tan \\theta }\\right)-\\tan \\theta \\left(\\tan \\theta \\right)} \\\\ &=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta } \\end{align}[\/latex]<\/p>\r\nWhen using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Verify the identity: [latex]\\cos \\left(2\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex].[reveal-answer q=\"584630\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584630\"][latex]\\cos \\left(2\\theta \\right)\\cos \\theta =\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex][\/hidden-answer]<\/section><\/div>","rendered":"

Using Double-Angle Formulas to Verify Identities<\/h2>\n

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.<\/p>\n

Establish the following identity using double-angle formulas:<\/p>\n

[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\n