{"id":1814,"date":"2025-07-28T20:27:46","date_gmt":"2025-07-28T20:27:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1814"},"modified":"2025-08-13T03:06:45","modified_gmt":"2025-08-13T03:06:45","slug":"the-other-trigonometric-functions-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/the-other-trigonometric-functions-learn-it-5\/","title":{"raw":"The Other Trigonometric Functions: Learn It 5","rendered":"The Other Trigonometric Functions: Learn It 5"},"content":{"raw":"

Alternate Forms of the Pythagorean Identity<\/h2>\r\nWe can use these fundamental identities to derive alternative forms of the Pythagorean Identity<\/strong>, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. One form is obtained by dividing both sides by [latex]{\\cos }^{2}t:[\/latex]\r\n
[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\cos }^{2}t}+\\frac{{\\sin }^{2}t}{{\\cos }^{2}t}=\\frac{1}{{\\cos }^{2}t}\\\\ \\\\ 1+{\\tan }^{2}t={\\sec }^{2}t\\end{gathered}[\/latex]<\/div>\r\nThe other form is obtained by dividing both sides by [latex]{\\sin }^{2}t:[\/latex]\r\n
[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\sin }^{2}t}+\\frac{{\\sin }^{2}t}{{\\sin }^{2}t}=\\frac{1}{{\\sin }^{2}t}\\\\ \\\\ {\\cot }^{2}t+1={\\csc }^{2}t\\end{gathered}[\/latex]<\/div>\r\n
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alternate Pythagorean identities<\/h3>\r\n

[latex]1+{\\tan }^{2}t={\\sec }^{2}t[\/latex]<\/p>\r\n

[latex]{\\cot }^{2}t+1={\\csc }^{2}t[\/latex]<\/p>\r\n\r\n<\/section><\/div>\r\n

If [latex]\\text{cos}\\left(t\\right)=\\frac{12}{13}[\/latex] and [latex]t[\/latex] is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 8<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"627612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"627612\"]\r\n\r\nWe can find the sine using the Pythagorean Identity, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex], and the remaining functions by relating them to sine and cosine.\r\n

[latex]\\begin{gathered}{\\left(\\frac{12}{13}\\right)}^{2}+{\\sin }^{2}t=1 \\\\ {\\sin }^{2}t=1-{\\left(\\frac{12}{13}\\right)}^{2} \\\\ {\\sin }^{2}t=1-\\frac{144}{169} \\\\ {\\sin }^{2}t=\\frac{25}{169} \\\\ \\sin t=\\pm \\sqrt{\\frac{25}{169}} \\\\ \\sin t=\\pm \\frac{\\sqrt{25}}{\\sqrt{169}} \\\\ \\sin t=\\pm \\frac{5}{13} \\end{gathered}[\/latex]<\/p>\r\nThe sign of the sine depends on the y<\/em>-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y<\/em>-values are negative, its sine is negative, [latex]-\\frac{5}{13}[\/latex].\r\n\r\nThe remaining functions can be calculated using identities relating them to sine and cosine.\r\n

[latex]\\begin{gathered}\\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{5}{13}}{\\frac{12}{13}}=-\\frac{5}{12} \\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{12}{13}}=\\frac{13}{12} \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{5}{13}}=-\\frac{13}{5}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\frac{5}{12}}=-\\frac{12}{5}\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

If [latex]\\sec \\left(t\\right)=-\\frac{17}{8}[\/latex] and [latex]0<t<\\pi [\/latex], find the values of the other five functions.[reveal-answer q=\"568112\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568112\"][latex]\\cos t=-\\frac{8}{17},\\sin t=\\frac{15}{17},\\tan t=-\\frac{15}{8},\\csc t=\\frac{17}{15},\\cot t=-\\frac{8}{15}[\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]100893[\/ohm_question]<\/section>
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periodic functions<\/h3>\r\nA function that repeats its values in regular intervals is known as a periodic function<\/strong>. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or [latex]2\\pi [\/latex], will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.\r\n\r\n<\/section>Other functions can also be periodic. For example, the lengths of months repeat every four years. If [latex]x[\/latex] represents the length time, measured in years, and [latex]f\\left(x\\right)[\/latex] represents the number of days in February, then [latex]f\\left(x+4\\right)=f\\left(x\\right)[\/latex].\u00a0This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that satisfies this condition and is called the period. A period<\/strong> is the shortest interval over which a function completes one full cycle\u2014in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days.\r\n\r\n
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period of a function<\/h3>\r\nThe period<\/strong> [latex]P[\/latex] of a repeating function [latex]f[\/latex] is the number representing the interval such that [latex]f\\left(x+P\\right)=f\\left(x\\right)[\/latex] for any value of [latex]x[\/latex].\r\n\r\nThe period of the cosine, sine, secant, and cosecant functions is [latex]2\\pi [\/latex].\r\n\r\nThe period of the tangent and cotangent functions is [latex]\\pi [\/latex].\r\n\r\n<\/section>
Find the values of the six trigonometric functions of angle [latex]t[\/latex] based on Figure 9.<\/strong>\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 9<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"99481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99481\"]\r\n

[latex]\\begin{gathered}\\sin t=y=-\\frac{\\sqrt{3}}{2}\\\\ \\cos t=x=-\\frac{1}{2}\\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{-\\frac{1}{2}}=\\sqrt{3}\\\\ \\sec t=\\frac{1}{\\cos t}=\\frac{1}{-\\frac{1}{2}}=-2\\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}=-\\frac{2\\sqrt{3}}{3}\\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n
\r\n\r\nFind the values of the six trigonometric functions of angle [latex]t[\/latex]\u00a0based on Figure 10.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 10<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"766306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"766306\"]\r\n\r\n[latex]\\begin{align}&\\sin t=-1\\\\&\\cos t=0\\\\&\\tan t \\text{ is undefined}\\\\ &\\sec t \\text{ is undefined}\\\\&\\csc t=-1\\\\&\\cot t=0\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
If [latex]\\sin \\left(t\\right)=-\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\\text{cos}\\left(t\\right)=\\frac{1}{2}[\/latex], find [latex]\\text{sec}\\left(t\\right),\\text{csc}\\left(t\\right),\\text{tan}\\left(t\\right),\\text{ cot}\\left(t\\right)[\/latex].[reveal-answer q=\"940244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"940244\"]\r\n

[latex]\\begin{gathered} \\sec t=\\frac{1}{\\cos t}=\\frac{1}{\\frac{1}{2}}=2 \\\\ \\csc t=\\frac{1}{\\sin t}=\\frac{1}{-\\frac{\\sqrt{3}}{2}}-\\frac{2\\sqrt{3}}{3} \\\\ \\tan t=\\frac{\\sin t}{\\cos t}=\\frac{-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=-\\sqrt{3} \\\\ \\cot t=\\frac{1}{\\tan t}=\\frac{1}{-\\sqrt{3}}=-\\frac{\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

If [latex]\\sin \\left(t\\right)=\\frac{\\sqrt{2}}{2}[\/latex]\u00a0and [latex]\\cos \\left(t\\right)=\\frac{\\sqrt{2}}{2}[\/latex], find [latex]\\text{sec}\\left(t\\right),\\text{csc}\\left(t\\right),\\text{tan}\\left(t\\right),\\text{ and cot}\\left(t\\right)[\/latex].[reveal-answer q=\"701998\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"701998\"][latex]\\sec t=\\sqrt{2},\\csc t=\\sqrt{2},\\tan t=1,\\cot t=1[\/latex][\/hidden-answer]<\/section>","rendered":"

Alternate Forms of the Pythagorean Identity<\/h2>\n

We can use these fundamental identities to derive alternative forms of the Pythagorean Identity<\/strong>, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. One form is obtained by dividing both sides by [latex]{\\cos }^{2}t:[\/latex]<\/p>\n

[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\cos }^{2}t}+\\frac{{\\sin }^{2}t}{{\\cos }^{2}t}=\\frac{1}{{\\cos }^{2}t}\\\\ \\\\ 1+{\\tan }^{2}t={\\sec }^{2}t\\end{gathered}[\/latex]<\/div>\n

The other form is obtained by dividing both sides by [latex]{\\sin }^{2}t:[\/latex]<\/p>\n

[latex]\\begin{gathered}\\frac{{\\cos }^{2}t}{{\\sin }^{2}t}+\\frac{{\\sin }^{2}t}{{\\sin }^{2}t}=\\frac{1}{{\\sin }^{2}t}\\\\ \\\\ {\\cot }^{2}t+1={\\csc }^{2}t\\end{gathered}[\/latex]<\/div>\n
\n
\n

alternate Pythagorean identities<\/h3>\n

[latex]1+{\\tan }^{2}t={\\sec }^{2}t[\/latex]<\/p>\n

[latex]{\\cot }^{2}t+1={\\csc }^{2}t[\/latex]<\/p>\n<\/section>\n<\/div>\n

If [latex]\\text{cos}\\left(t\\right)=\\frac{12}{13}[\/latex] and [latex]t[\/latex] is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.
\n<\/span><\/p>\n
\"Graph
Figure 8<\/b><\/figcaption><\/figure>\n