{"id":1995,"date":"2025-07-31T23:28:30","date_gmt":"2025-07-31T23:28:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1995"},"modified":"2025-10-14T19:09:39","modified_gmt":"2025-10-14T19:09:39","slug":"double-angle-half-angle-and-reduction-formulas-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas-learn-it-4\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas: Learn It 4","rendered":"Double Angle, Half Angle, and Reduction Formulas: Learn It 4"},"content":{"raw":"

Using Half-Angle Formulas to Find Exact Values<\/h2>\r\nThe next set of identities is the set of half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta [\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm [\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.\r\n\r\nThe half-angle formula for sine is derived as follows:\r\n
[latex]\\begin{align}{\\sin }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nTo derive the half-angle formula for cosine, we have\r\n
[latex]\\begin{align}{\\cos }^{2}\\theta &=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nFor the tangent identity, we have\r\n
[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\r\n
\r\n

half angle formulas<\/h3>\r\n[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]\r\n\r\n<\/section><\/div>\r\n
Find [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.[reveal-answer q=\"283155\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283155\"]Since [latex]{15}^{\\circ }=\\frac{{30}^{\\circ }}{2}[\/latex], we use the half-angle formula for sine:\r\n

[latex]\\begin{align} \\sin \\frac{{30}^{\\circ }}{2}&=\\sqrt{\\frac{1-\\cos {30}^{\\circ }}{2}} \\\\ &=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{2-\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{2-\\sqrt{3}}{4}} \\\\ &=\\frac{\\sqrt{2-\\sqrt{3}}}{2} \\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nNotice that we used only the positive root because [latex]\\sin \\left({15}^{\\text{o}}\\right)[\/latex] is positive.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.<\/strong>\r\n
    \r\n \t
  1. Draw a triangle to represent the given information.<\/li>\r\n \t
  2. Determine the correct half-angle formula.<\/li>\r\n \t
  3. Substitute values into the formula based on the triangle.<\/li>\r\n \t
  4. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>
    Given that [latex]\\tan \\alpha =\\frac{8}{15}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant III, find the exact value of the following:\r\n
      \r\n \t
    1. [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t
    2. [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t
    3. [latex]\\tan \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"751659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"751659\"]\r\n\r\nUsing the given information, we can draw the triangle. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\\sin \\alpha =-\\frac{8}{17}[\/latex] and [latex]\\cos \\alpha =-\\frac{15}{17}[\/latex].\r\n\r\n\"Diagram\r\n
        \r\n \t
      1. Before we start, we must remember that, if [latex]\\alpha [\/latex] is in quadrant III, then [latex]180^\\circ <\\alpha <270^\\circ [\/latex], so [latex]\\frac{180^\\circ }{2}<\\frac{\\alpha }{2}<\\frac{270^\\circ }{2}[\/latex]. This means that the terminal side of [latex]\\frac{\\alpha }{2}[\/latex] is in quadrant II, since [latex]90^\\circ <\\frac{\\alpha }{2}<135^\\circ [\/latex].To find [latex]\\sin \\frac{\\alpha }{2}[\/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle and simplify.\r\n
        [latex]\\begin{align} \\sin \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{32}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{16}{17}} \\\\ &=\\pm \\frac{4}{\\sqrt{17}} \\\\ &=\\frac{4\\sqrt{17}}{17} \\end{align}[\/latex]<\/div>\r\nWe choose the positive value of [latex]\\sin \\frac{\\alpha }{2}[\/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.<\/li>\r\n \t
      2. To find [latex]\\cos \\frac{\\alpha }{2}[\/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle, and simplify.\r\n
        [latex]\\begin{align} \\cos \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1+\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{2}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{2}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{1}{17}} \\\\ &=-\\frac{\\sqrt{17}}{17}\\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\cos \\frac{\\alpha }{2}[\/latex] because the angle is in quadrant II because cosine is negative in quadrant II.<\/li>\r\n \t
      3. To find [latex]\\tan \\frac{\\alpha }{2}[\/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle and simplify.\r\n
        [latex]\\begin{align} \\tan \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{1+\\left(-\\frac{15}{17}\\right)}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{\\frac{2}{17}}} \\\\ &=\\pm \\sqrt{\\frac{32}{2}} \\\\ &=-\\sqrt{16} \\\\ &=-4 \\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\tan \\frac{\\alpha }{2}[\/latex] because [latex]\\frac{\\alpha }{2}[\/latex] lies in quadrant II, and tangent is negative in quadrant II.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
        Given that [latex]\\sin \\alpha =-\\frac{4}{5}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant IV, find the exact value of [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex].[reveal-answer q=\"406030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406030\"][latex]-\\frac{2}{\\sqrt{5}}[\/latex][\/hidden-answer]<\/section>
        [ohm_question hide_question_numbers=1]173569[\/ohm_question]<\/section><\/div>","rendered":"

        Using Half-Angle Formulas to Find Exact Values<\/h2>\n

        The next set of identities is the set of half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta[\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm[\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.<\/p>\n

        The half-angle formula for sine is derived as follows:<\/p>\n

        [latex]\\begin{align}{\\sin }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n

        To derive the half-angle formula for cosine, we have<\/p>\n

        [latex]\\begin{align}{\\cos }^{2}\\theta &=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n

        For the tangent identity, we have<\/p>\n

        [latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\n
        \n
        \n

        half angle formulas<\/h3>\n

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

        Find [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.<\/p>\n