{"id":221009,"date":"2025-05-29T05:47:53","date_gmt":"2025-05-29T05:47:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221009"},"modified":"2025-05-29T05:47:56","modified_gmt":"2025-05-29T05:47:56","slug":"find-the-terminal-point-determined","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/find-the-terminal-point-determined\/","title":{"rendered":"Find The Terminal Point Determined"},"content":{"rendered":"\n

Consider The Following. T = – 5pi\/4 (A) Find The Reference Number T For The Value Of T. T = (B) Find The Terminal Point Determined By T. (X, Y) = (_<\/em><\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

\n\n\n\n

(A) Find the Reference Number for<\/strong><\/h3>\n\n\n\n

T=\u22125\u03c04T = -\\frac{5\\pi}{4}<\/strong><\/p>\n\n\n\n

Step 1: Find the coterminal angle between 00 and 2\u03c02\\pi<\/strong>
To do this, we add 2\u03c02\\pi until the angle is in the interval [0,2\u03c0)[0, 2\\pi): \u22125\u03c04+2\u03c0=\u22125\u03c04+8\u03c04=3\u03c04-\\frac{5\\pi}{4} + 2\\pi = -\\frac{5\\pi}{4} + \\frac{8\\pi}{4} = \\frac{3\\pi}{4}<\/p>\n\n\n\n

So, the coterminal angle is 3\u03c04\\frac{3\\pi}{4}, which lies in Quadrant II<\/strong>.<\/p>\n\n\n\n

Step 2: Find the reference number<\/strong><\/p>\n\n\n\n

In Quadrant II<\/strong>, the reference number is: \u03c0\u22123\u03c04=\u03c04\\pi – \\frac{3\\pi}{4} = \\frac{\\pi}{4}<\/p>\n\n\n\n

\u2705 Answer (A):<\/strong>
Reference Number = \u03c04\\frac{\\pi}{4}<\/strong><\/p>\n\n\n\n

\n\n\n\n

(B) Find the Terminal Point Determined by<\/strong><\/h3>\n\n\n\n

T=\u22125\u03c04T = -\\frac{5\\pi}{4}<\/strong><\/p>\n\n\n\n

Since all points on the unit circle are of the form (cos\u2061T,sin\u2061T)(\\cos T, \\sin T), we compute the cosine and sine of \u22125\u03c04-\\frac{5\\pi}{4}.<\/p>\n\n\n\n

Alternatively, we can recognize that \u22125\u03c04-\\frac{5\\pi}{4} is coterminal with 3\u03c04\\frac{3\\pi}{4}, but since signs differ depending on quadrant, we use the original<\/strong> angle\u2019s quadrant.<\/p>\n\n\n\n

    \n
  • T=\u22125\u03c04T = -\\frac{5\\pi}{4} lies in Quadrant III<\/strong> (moving clockwise from 0).<\/li>\n\n\n\n
  • In Quadrant III, both cosine and sine are negative<\/strong>.<\/li>\n<\/ul>\n\n\n\n

    From the reference number \u03c04\\frac{\\pi}{4}, we know: cos\u2061(\u03c04)=sin\u2061(\u03c04)=22\\cos\\left(\\frac{\\pi}{4}\\right) = \\sin\\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}<\/p>\n\n\n\n

    Thus, the coordinates for T=\u22125\u03c04T = -\\frac{5\\pi}{4} are: (x,y)=(\u221222,\u221222)(x, y) = \\left(-\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)<\/p>\n\n\n\n

    \u2705 Answer (B):<\/strong>
    Terminal Point = (\u221222,\u221222)\\left(-\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    \ud83d\udcd8 Explanation<\/h3>\n\n\n\n

    To solve trigonometric problems involving angles on the unit circle, two key concepts are used: reference number<\/strong> and terminal point<\/strong>. The reference number helps simplify the angle into an acute version (between 0 and \u03c02\\frac{\\pi}{2}) within the first quadrant, where trigonometric values are easiest to evaluate.<\/p>\n\n\n\n

    In part (A), the original angle is T=\u22125\u03c04T = -\\frac{5\\pi}{4}, a negative angle, meaning it rotates clockwise<\/strong> from the positive x-axis. To bring this into a more usable form, we find a coterminal angle<\/strong> by adding 2\u03c02\\pi (a full circle): \u22125\u03c04+2\u03c0=3\u03c04-\\frac{5\\pi}{4} + 2\\pi = \\frac{3\\pi}{4}<\/p>\n\n\n\n

    Now in standard position, 3\u03c04\\frac{3\\pi}{4} is in Quadrant II<\/strong>. The reference number is the acute angle between the terminal side and the x-axis. In Quadrant II, we subtract the angle from \u03c0\\pi: \u03c0\u22123\u03c04=\u03c04\\pi – \\frac{3\\pi}{4} = \\frac{\\pi}{4}<\/p>\n\n\n\n

    This gives us the reference number: \u03c04\\frac{\\pi}{4}.<\/p>\n\n\n\n

    In part (B), we identify which quadrant \u22125\u03c04-\\frac{5\\pi}{4} falls in. Moving clockwise past \u2212\u03c0-\\pi (or \u22124\u03c0\/4-4\\pi\/4), we land in Quadrant III<\/strong>. Here, both sine and cosine are negative<\/strong>. The terminal point uses the same values as \u03c04\\frac{\\pi}{4} (since that\u2019s the reference angle), but with the appropriate signs: (cos\u2061T,sin\u2061T)=(\u221222,\u221222)(\\cos T, \\sin T) = \\left(-\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)<\/p>\n\n\n\n

    This point lies on the unit circle and represents the exact position of the angle \u22125\u03c04-\\frac{5\\pi}{4}.<\/p>\n\n\n\n

    \n\n\n\n

    Let me know if you’d like a diagram or visual explanation!<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Consider The Following. T = – 5pi\/4 (A) Find The Reference Number T For The Value Of T. T = (B) Find The Terminal Point Determined By T. (X, Y) = (_ The Correct Answer and Explanation is: (A) Find the Reference Number for T=\u22125\u03c04T = -\\frac{5\\pi}{4} Step 1: Find the coterminal angle between 00 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-221009","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221009","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=221009"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221009\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221009"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221009"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}