{"id":9504,"date":"2017-03-28T09:57:19","date_gmt":"2017-03-28T16:57:19","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=9504"},"modified":"2022-06-14T09:15:49","modified_gmt":"2022-06-14T16:15:49","slug":"ap-calculus-review-trigonometric-identities","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-review-trigonometric-identities\/","title":{"rendered":"AP Calculus Review: Trigonometric Identities"},"content":{"rendered":"<p>What are trigonometric identities doing on a Calculus exam? I&#8217;m glad you asked! Because mathematics tends to build upon previous results, you can expect to use all of your prior math knowledge when working on calculus problems.  <\/p>\n<p>Trigonometry is useful when setting up problems involving right triangles.  Moreover, the trigonometric identities also help when working out limits, derivatives and integrals of trig functions.  Specifically, these identities seem to come up more often when working out integrals, especially on the no-calculator sections of the test.<\/p>\n<h2>The Unit Circle<\/h2>\n<p>First of all, you should be familiar with the <strong>Unit Circle<\/strong>.  Each point (<em>x<\/em>, <em>y<\/em>) on the circle gives the value of cosine and sine of the corresponding angle.<\/p>\n<p>For example, look for the angle 2&pi;\/3 (radians).  Based on the unit circle diagram, you know:<\/p>\n<ul>\n<li>cos 2&pi;\/3 = -1\/2.<\/li>\n<li>sin 2&pi;\/3 = &radic;<span style=\"text-decoration:overline\">&nbsp;3&nbsp;<\/span>\/2.\n<\/li>\n<\/ul>\n<figure id=\"attachment_9507\" aria-describedby=\"caption-attachment-9507\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_-600x600.png\" alt=\"Unit circle - trigonometric identities\" width=\"600\" height=\"600\" class=\"size-large wp-image-9507\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_-600x600.png 600w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_-300x300.png 300w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_-150x150.png 150w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_-768x768.png 768w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/1024px-Unit_Circle_Angles_Color_Clock_Face.svg_.png 1024w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><figcaption id=\"caption-attachment-9507\" class=\"wp-caption-text\">The Unit Circle, which displays cosine and sine values of common angles.<\/figcaption><\/figure>\n<h2>The Essential Trigonometric Identities<\/h2>\n<p>Fortunately, you do not have to remember absolutely every identity from Trig class.  Below is a list of what I would consider the essential identities.<\/p>\n<h3>1. Quotient Identities<\/h3>\n<p>The quotient identities are useful for re-expressing the trig functions in terms of sin and\/or cos.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/quotient_identities-600x52.png\" alt=\"Quotient identities\" width=\"600\" height=\"52\" class=\"aligncenter size-large wp-image-9508\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/quotient_identities-600x52.png 600w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/quotient_identities-300x26.png 300w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/quotient_identities.png 628w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><\/p>\n<h3>2. Even and Odd Properties<\/h3>\n<p>The following rules provide shortcuts for dealing with negative input.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_symmetry.png\" alt=\"Trig symmetry properties\" width=\"512\" height=\"42\" class=\"aligncenter size-full wp-image-9509\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_symmetry.png 512w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_symmetry-300x25.png 300w\" sizes=\"(max-width: 512px) 100vw, 512px\" \/><\/p>\n<h3>3. Pythagorean Identities<\/h3>\n<p>Perhaps the most famous and useful equation in all of mathematics is the Pythagorean Theorem.  If the side lengths of a right triangle are <em>a<\/em> and <em>b<\/em>, and the hypotenuse is <em>c<\/em>, then:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_Theorem.gif\" alt=\"Pythagorean Theorem\" width=\"92\" height=\"17\" class=\"aligncenter size-full wp-image-9510\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_Theorem.gif 92w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_Theorem-30x6.gif 30w\" sizes=\"(max-width: 92px) 100vw, 92px\" \/><\/p>\n<p>When applied to the unit circle, we get a trio of useful trigonometric identities:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_identities.png\" alt=\"Pythagorean_identities\" width=\"544\" height=\"37\" class=\"aligncenter size-full wp-image-9511\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_identities.png 544w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_identities-300x20.png 300w\" sizes=\"(max-width: 544px) 100vw, 544px\" \/><\/p>\n<h3>4. Half-Angle Identities<\/h3>\n<p>The half-angle identities can be used to convert a squared expression into a form that is easier to deal with.  So if sine-squared or cosine-squared shows up in an integral, remember these identities.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Half-angle_identites-600x60.png\" alt=\"Half-angle identities\" width=\"600\" height=\"60\" class=\"aligncenter size-large wp-image-9513\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Half-angle_identites-600x60.png 600w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Half-angle_identites-300x30.png 300w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Half-angle_identites.png 616w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><\/p>\n<h2>Examples<\/h2>\n<p>As you can see, the list of essential trig identities is not terribly long.  In fact, there are just a few key items to remember.  Now let&#8217;s see how we can use these identities in actual AP Calculus problems.<\/p>\n<h3>Using the Unit Circle<\/h3>\n<p>Find the area of the shaded region between <em>y<\/em> = sin <em>x<\/em> and the line <em>y<\/em> = 1\/2.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/graph_sin_and_line.png\" alt=\"Graph of y = sin x and line y = 1\/2\" width=\"300\" height=\"300\" class=\"aligncenter size-full wp-image-9514\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/graph_sin_and_line.png 300w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/graph_sin_and_line-150x150.png 150w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>We know that a properly set-up integral will do the trick.  But what should the limits of integration be?  We first have to solve sin <em>x<\/em> = 1\/2.<\/p>\n<p>Based on the unit circle, there are two angles at which the sine value is 1\/2: &nbsp;&nbsp; &pi;\/6 and 5&pi;\/6.  Therefore, the area of the shaded region is:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_solution1.gif\" alt=\"Solution of Example 1, first part\" width=\"398\" height=\"54\" class=\"aligncenter size-full wp-image-9515\" \/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_solution2.gif\" alt=\"Part two of solution\" width=\"596\" height=\"55\" class=\"aligncenter size-full wp-image-9516\" \/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_solution3.gif\" alt=\"Final solution to Example 1\" width=\"185\" height=\"42\" class=\"aligncenter size-full wp-image-9517\" \/><\/p>\n<p>Note how we also used the unit circle at the end to determine the values of cos &pi;\/6 and 5cos &pi;\/6.<\/p>\n<h3>Using the Quotient and Pythagorean Identities<\/h3>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral.gif\" alt=\"A difficult trig integral\" width=\"203\" height=\"45\" class=\"aligncenter size-full wp-image-9518\" \/><\/p>\n<h4>Solution<\/h4>\n<p>At first, this integral seems impossible!  However with a bit of skillful manipulation, this one transforms into an easier problem.<\/p>\n<p>First let&#8217;s re-express the division by cos<sup>2<\/sup> <em>x<\/em> as a multiplication by its reciprocal, sec<sup>2<\/sup> <em>x<\/em>.  <\/p>\n<p>Then notice that an additional sec<sup>2<\/sup> <em>x<\/em> can be factored out.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral_solution_1.gif\" alt=\"Part 1 of solution to trig integral\" width=\"283\" height=\"88\" class=\"aligncenter size-full wp-image-9519\" \/><\/p>\n<p>Next, we use a form of the Pythagorean identity.  <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Pythagorean_identity.gif\" alt=\"Rewriting the Pythagorean Identity\" width=\"333\" height=\"17\" class=\"aligncenter size-full wp-image-9520\" \/><\/p>\n<p>Therefore, we can rewrite the integral as:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral_solution_2.gif\" alt=\"Second part of solution\" width=\"339\" height=\"41\" class=\"aligncenter size-full wp-image-9521\" \/><\/p>\n<p>Notice the cosines and secants.  As we know, those two trig functions are reciprocals of one another.  So, just like 4 and 1\/4 would cancel when multiplied together, we can cancel two of the cosines and secants.  This leaves a very simple integral:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral_solution_3.gif\" alt=\"Last part of the solution\" width=\"208\" height=\"41\" class=\"aligncenter size-full wp-image-9522\" \/><\/p>\n<p>It may seem strange that after all of that work, the final answer is so trivial.  But this is often the case when using trigonometric identities to manipulate the problem.  Test makers are really good at creating problems that unravel like this.<\/p>\n<h3>Using Even and Odd Properties<\/h3>\n<p>Describe the vector function (cos(-<em>t<\/em>), sin(-<em>t<\/em>)), where 0 &le; <em>t<\/em> &le; 2&pi;.<\/p>\n<p>(A) A circle traced counterclockwise starting and ending at (1, 0)<\/p>\n<p>(B) A circle traced clockwise starting and ending at (1, 0)<\/p>\n<p>(C) A circle traced counterclockwise starting and ending at (0, 1)<\/p>\n<p>(D) A circle traced clockwise starting and ending at (0, 1)<\/p>\n<h4>Solution<\/h4>\n<p>First, plug in <em>t<\/em> = 0.<\/p>\n<p>(cos(-0), sin(-0)) = (cos(0), sin(0)) = (1, 0).<\/p>\n<p>This rules out (C) and (D) immediately.<\/p>\n<p>Next, the easiest way to determine the correct direction is to use the even and odd properties of sine and cosine.  <\/p>\n<p>cos(-<em>t<\/em>) = cos(<em>t<\/em>).<\/p>\n<p>sin(-<em>t<\/em>) = -sin(<em>t<\/em>).<\/p>\n<p>Therefore, the original vector function is equivalent to: (cos(<em>t<\/em>), -sin(<em>t<\/em>)).  But this new path is simply a reflection of the original path over the <em>x<\/em>-axis, because the <em>y<\/em>-coordinate has been negated.<\/p>\n<p>If you recall that (cos(<em>t<\/em>), sin(<em>t<\/em>)) traces out the standard unit circle counterclockwise, then (cos(<em>t<\/em>), -sin(<em>t<\/em>)) traces it out the other way, that is, clockwise.<\/p>\n<p>The correct response is (B).<\/p>\n<h3>Using the Half-Angle Identities<\/h3>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_trig_integral.gif\" alt=\"Example.  The integral of sine-squared of 4x.\" width=\"107\" height=\"41\" class=\"aligncenter size-full wp-image-9523\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_trig_integral.gif 107w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/Example_trig_integral-30x11.gif 30w\" sizes=\"(max-width: 107px) 100vw, 107px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>The only way to proceed on this one is to use a half-angle identity.  Just be careful with the 4<em>x<\/em> inside.  You&#8217;ll have to multiply that by 2 in the formula.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral_solution_A.gif\" alt=\"First part of solution to the third example.\" width=\"342\" height=\"91\" class=\"aligncenter size-full wp-image-9524\" \/><\/p>\n<p>Let&#8217;s split up the work now. The first term is trivially easy.  <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/integral_of_1.gif\" alt=\"The integral of 1 is equal to x.\" width=\"124\" height=\"41\" class=\"aligncenter size-full wp-image-9525\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/integral_of_1.gif 124w, https:\/\/magoosh.com\/hs\/files\/2017\/03\/integral_of_1-30x10.gif 30w\" sizes=\"(max-width: 124px) 100vw, 124px\" \/><\/p>\n<p>But the second term requires a substitution.  Let <em>u<\/em> = 8<em>x<\/em>.  Then <em>du<\/em> = 8 <em>dx<\/em>, or equivalently, <em>dx<\/em> = (1\/8) <em>du<\/em>.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/integral_of_cos8x.gif\" alt=\"Integral of cos(8x) worked out\" width=\"236\" height=\"126\" class=\"aligncenter size-full wp-image-9526\" \/><\/p>\n<p>Putting these two results together with the main problem, we get:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/03\/Trig_integral_solution_B.gif\" alt=\"Final solution of the example.\" width=\"305\" height=\"88\" class=\"aligncenter size-full wp-image-9527\" \/><\/p>\n<h2>Summary<\/h2>\n<p>There are only a handful of trigonometric identities that you must master for the AP Calculus exams.  Together with knowing your way around the unit circle, make sure that you understand when and how to use:<\/p>\n<ol>\n<li>Quotient Identities<\/li>\n<li>Even and Odd Properties<\/li>\n<li>Pythagorean Identities, and<\/li>\n<li>Half-Angle Identities<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Trigonometric identities are used in problems involving triangles, but also when working out limits, derivatives and integrals of trigonometric functions.<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-9504","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>AP Calculus Review: Trigonometric Identities - 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