{"id":1971,"date":"2025-07-31T21:41:38","date_gmt":"2025-07-31T21:41:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1971"},"modified":"2025-08-13T03:12:15","modified_gmt":"2025-08-13T03:12:15","slug":"simplifying-trigonometric-expressions-with-identities-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/simplifying-trigonometric-expressions-with-identities-learn-it-2\/","title":{"raw":"Simplifying Trigonometric Expressions With Identities: Learn It 2","rendered":"Simplifying Trigonometric Expressions With Identities: Learn It 2"},"content":{"raw":"

Reciprocal and Quotient Identities<\/h2>\r\nThe next set of fundamental identities is the set of reciprocal identities<\/strong>, which, as their name implies, relate trigonometric functions that are reciprocals of each other.\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Reciprocal Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]\\sin \\theta =\\frac{1}{\\csc \\theta }[\/latex]<\/td>\r\n[latex]\\csc \\theta =\\frac{1}{\\sin \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\cos \\theta =\\frac{1}{\\sec \\theta }[\/latex]<\/td>\r\n[latex]\\sec \\theta =\\frac{1}{\\cos \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]\\tan \\theta =\\frac{1}{\\cot \\theta }[\/latex]<\/td>\r\n[latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe final set of identities is the set of quotient identities<\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n
Quotient Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]\\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }[\/latex]<\/td>\r\n[latex]\\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.\r\n\r\n
\r\n

trigonometric identities<\/h3>\r\nThe Pythagorean identities<\/strong> are based on the properties of a right triangle.\r\n

[latex]\\begin{gathered} {\\cos}^{2}\\theta + {\\sin}^{2}\\theta=1 \\\\ 1+{\\tan}^{2}\\theta={\\sec}^{2}\\theta \\\\ 1+{\\cot}^{2}\\theta={\\csc}^{2}\\theta\\end{gathered}[\/latex]<\/p>\r\nThe even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.\r\n

[latex]\\begin{gathered} \\cos(-\\theta)=\\cos(\\theta) \\\\\\sin(-\\theta)=-\\sin(\\theta) \\\\\\tan(-\\theta)=-\\tan(\\theta) \\\\\\cot(-\\theta)=-\\cot(\\theta) \\\\\\sec(-\\theta)=\\sec(\\theta) \\\\\\csc(-\\theta)=-\\csc(\\theta) \\end{gathered}[\/latex]<\/p>\r\nThe reciprocal identities<\/strong> define reciprocals of the trigonometric functions.\r\n

[latex]\\begin{gathered}\\sin\\theta=\\frac{1}{\\csc\\theta} \\\\ \\cos\\theta=\\frac{1}{\\sec\\theta} \\\\ \\tan\\theta=\\frac{1}{\\cot\\theta} \\\\ \\cot\\theta=\\frac{1}{\\tan\\theta} \\\\ \\sec\\theta=\\frac{1}{\\cos\\theta} \\\\ \\csc\\theta=\\frac{1}{\\sin\\theta}\\end{gathered}[\/latex]<\/p>\r\nThe quotient identities<\/strong> define the relationship among the trigonometric functions.\r\n

[latex]\\begin{gathered} \\tan\\theta=\\frac{\\sin\\theta}{\\cos\\theta} \\\\ \\cot\\theta=\\frac{\\cos\\theta}{\\sin\\theta} \\end{gathered}[\/latex]<\/p>\r\n\r\n<\/section>

Graph both sides of the identity [latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]. In other words, on the graphing calculator, graph [latex]y=\\cot \\theta[\/latex] and [latex]y=\\frac{1}{\\tan \\theta }[\/latex].[reveal-answer q=\"943922\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943922\"]\r\n

\"Graph<\/span><\/h3>\r\n \r\n

Analysis of the Solution<\/h4>\r\nWe see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Verify [latex]\\tan \\theta \\cos \\theta =\\sin \\theta[\/latex].[reveal-answer q=\"136260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"136260\"]We will start on the left side, as it is the more complicated side:\r\n

[latex]\\begin{align}\\tan \\theta \\cos \\theta &=\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\cos \\theta \\\\ &=\\left(\\frac{\\sin \\theta }{\\cancel{\\cos \\theta }}\\right)\\cancel{\\cos \\theta } \\\\ &=\\sin \\theta \\end{align}[\/latex]<\/p>\r\n\r\n

Analysis of the Solution<\/h4>\r\nThis identity was fairly simple to verify, as it only required writing [latex]\\tan \\theta[\/latex] in terms of [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
Verify the identity [latex]\\csc \\theta \\cos \\theta \\tan \\theta =1[\/latex].[reveal-answer q=\"361361\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"361361\"]\r\n

[latex]\\begin{align}\\csc \\theta \\cos \\theta \\tan \\theta &=\\left(\\frac{1}{\\sin \\theta }\\right)\\cos \\theta \\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &=\\frac{\\cos \\theta }{\\sin \\theta }\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &=\\frac{\\sin \\theta \\cos \\theta }{\\sin \\theta \\cos \\theta } \\\\ &=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

\r\n

Even and Odd Identities<\/h3>\r\n\r\n\r\n\r\n\r\n\r\n
Even-Odd Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]\\begin{gathered}\\tan \\left(-\\theta \\right)=-\\tan \\theta\\\\ \\cot \\left(-\\theta \\right)=-\\cot \\theta \\end{gathered}[\/latex]<\/td>\r\n[latex]\\begin{gathered}\\sin \\left(-\\theta \\right)=-\\sin \\theta\\\\ \\csc \\left(-\\theta \\right)=-\\csc \\theta\\end{gathered}[\/latex]<\/td>\r\n[latex]\\begin{gathered}\\cos \\left(-\\theta \\right)=\\cos \\theta \\\\ \\sec \\left(-\\theta \\right)=\\sec \\theta \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.\r\n\r\n<\/div>\r\n
Verify the following equivalency using the even-odd identities:\r\n

[latex]\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]={\\cos }^{2}x[\/latex]<\/p>\r\n[reveal-answer q=\"208801\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"208801\"]\r\n\r\nWorking on the left side of the equation, we have\r\n

[latex]\\begin{align}\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]&=\\left(1+\\sin x\\right)\\left(1-\\sin x\\right)&& \\text{Since sin(-}x\\text{)=}-\\sin x \\\\ &=1-{\\sin }^{2}x&& \\text{Difference of squares} \\\\ &={\\cos }^{2}x&& {\\text{cos}}^{2}x=1-{\\sin }^{2}x\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n

Show that [latex]\\frac{\\cot \\theta }{\\csc \\theta }=\\cos \\theta[\/latex].[reveal-answer q=\"813945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"813945\"]\r\n

[latex]\\begin{align}\\frac{\\cot \\theta }{\\csc \\theta }&=\\frac{\\frac{\\cos \\theta }{\\sin \\theta }}{\\frac{1}{\\sin \\theta }} \\\\ &=\\frac{\\cos \\theta }{\\sin \\theta }\\cdot \\frac{\\sin \\theta }{1} \\\\ &=\\cos \\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n

Create an identity for the expression [latex]2\\tan \\theta \\sec \\theta[\/latex] by rewriting strictly in terms of sine.[reveal-answer q=\"482916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482916\"]There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:\r\n

[latex]\\begin{align}2\\tan \\theta \\sec \\theta &=2\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\left(\\frac{1}{\\cos \\theta }\\right) \\\\ &=\\frac{2\\sin \\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }&& \\text{Substitute }1-{\\sin }^{2}\\theta \\text{ for }{\\cos }^{2}\\theta \\end{align}[\/latex]<\/p>\r\nThus,\r\n

[latex]2\\tan \\theta \\sec \\theta =\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Verify the identity:\r\n

[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"689339\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689339\"]\r\n\r\nLet\u2019s start with the left side and simplify:\r\n

[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}&=\\frac{{\\left[\\sin \\left(-\\theta \\right)\\right]}^{2}-{\\left[\\cos \\left(-\\theta \\right)\\right]}^{2}}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)} \\\\ &=\\frac{{\\left(-\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&& \\sin \\left(-x\\right)=-\\sin x\\text{ and }\\cos \\left(-x\\right)=\\cos x \\\\ &=\\frac{{\\left(\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&& \\text{Difference of squares} \\\\ &=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\sin \\theta +\\cos \\theta \\right)}{-\\left(\\sin \\theta +\\cos \\theta \\right)} \\\\ &=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)}{-\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)} \\\\ &=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Verify the identity [latex]\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }=\\frac{\\sin \\theta +1}{\\tan \\theta }[\/latex].[reveal-answer q=\"307900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307900\"][latex]\\begin{align}\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }&=\\frac{\\left(\\sin \\theta +1\\right)\\left(\\sin \\theta -1\\right)}{\\tan \\theta \\left(\\sin \\theta -1\\right)}\\\\ &=\\frac{\\sin \\theta +1}{\\tan \\theta }\\end{align}[\/latex][\/hidden-answer]<\/section>
Verify the identity: [latex]\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)=1[\/latex].[reveal-answer q=\"690675\"]Show Solutions[\/reveal-answer]\r\n[hidden-answer a=\"690675\"]We will work on the left side of the equation.\r\n

[latex]\\begin{align}\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)&=\\left(1-{\\cos }^{2}x\\right)\\left(1+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x}{{\\sin }^{2}x}+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) && \\text{Find the common denominator}. \\\\ &=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x+{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &=\\left({\\sin }^{2}x\\right)\\left(\\frac{1}{{\\sin }^{2}x}\\right) \\\\ &=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Reciprocal and Quotient Identities<\/h2>\n

The next set of fundamental identities is the set of reciprocal identities<\/strong>, which, as their name implies, relate trigonometric functions that are reciprocals of each other.<\/p>\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n
Reciprocal Identities<\/th>\n<\/tr>\n<\/thead>\n
[latex]\\sin \\theta =\\frac{1}{\\csc \\theta }[\/latex]<\/td>\n[latex]\\csc \\theta =\\frac{1}{\\sin \\theta }[\/latex]<\/td>\n<\/tr>\n
[latex]\\cos \\theta =\\frac{1}{\\sec \\theta }[\/latex]<\/td>\n[latex]\\sec \\theta =\\frac{1}{\\cos \\theta }[\/latex]<\/td>\n<\/tr>\n
[latex]\\tan \\theta =\\frac{1}{\\cot \\theta }[\/latex]<\/td>\n[latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The final set of identities is the set of quotient identities<\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.<\/p>\n\n\n\n<\/colgroup>\n\n\n\n\n
Quotient Identities<\/th>\n<\/tr>\n<\/thead>\n
[latex]\\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }[\/latex]<\/td>\n[latex]\\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.<\/p>\n

\n

trigonometric identities<\/h3>\n

The Pythagorean identities<\/strong> are based on the properties of a right triangle.<\/p>\n

[latex]\\begin{gathered} {\\cos}^{2}\\theta + {\\sin}^{2}\\theta=1 \\\\ 1+{\\tan}^{2}\\theta={\\sec}^{2}\\theta \\\\ 1+{\\cot}^{2}\\theta={\\csc}^{2}\\theta\\end{gathered}[\/latex]<\/p>\n

The even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.<\/p>\n

[latex]\\begin{gathered} \\cos(-\\theta)=\\cos(\\theta) \\\\\\sin(-\\theta)=-\\sin(\\theta) \\\\\\tan(-\\theta)=-\\tan(\\theta) \\\\\\cot(-\\theta)=-\\cot(\\theta) \\\\\\sec(-\\theta)=\\sec(\\theta) \\\\\\csc(-\\theta)=-\\csc(\\theta) \\end{gathered}[\/latex]<\/p>\n

The reciprocal identities<\/strong> define reciprocals of the trigonometric functions.<\/p>\n

[latex]\\begin{gathered}\\sin\\theta=\\frac{1}{\\csc\\theta} \\\\ \\cos\\theta=\\frac{1}{\\sec\\theta} \\\\ \\tan\\theta=\\frac{1}{\\cot\\theta} \\\\ \\cot\\theta=\\frac{1}{\\tan\\theta} \\\\ \\sec\\theta=\\frac{1}{\\cos\\theta} \\\\ \\csc\\theta=\\frac{1}{\\sin\\theta}\\end{gathered}[\/latex]<\/p>\n

The quotient identities<\/strong> define the relationship among the trigonometric functions.<\/p>\n

[latex]\\begin{gathered} \\tan\\theta=\\frac{\\sin\\theta}{\\cos\\theta} \\\\ \\cot\\theta=\\frac{\\cos\\theta}{\\sin\\theta} \\end{gathered}[\/latex]<\/p>\n<\/section>\n

Graph both sides of the identity [latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]. In other words, on the graphing calculator, graph [latex]y=\\cot \\theta[\/latex] and [latex]y=\\frac{1}{\\tan \\theta }[\/latex].<\/p>\n