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{"id":1078,"date":"2022-08-22T00:10:10","date_gmt":"2022-08-22T00:10:10","guid":{"rendered":"https:\/\/mathwild.com\/?page_id=1078"},"modified":"2023-09-12T23:34:00","modified_gmt":"2023-09-12T23:34:00","slug":"tangent-multiple-angle-formula-for-sine-cosine","status":"publish","type":"page","link":"https:\/\/mathwild.com\/tangent-multiple-angle-formula-for-sine-cosine\/","title":{"rendered":"Tangent: Multiple Angle Formula for Sine\/Cosine"},"content":{"rendered":"\n

For those of you who tried to derive the binomial expansion formula yourselves, here’s the step-by-step process below, based on how I derived the result when first studying complex numbers last year:<\/p>\n\n\n\n

Setting \\( x=\\cos(\\theta), y=i\\sin(\\theta) \\) we can expand below:<\/p>\n\n\n\n

<\/div>\n\n\n\n
(\\cos(\\theta)+i\\sin(\\theta))^n=\\sum_{k=0}^{n}\\binom{n}{k}\\cos^{n-k}(\\theta)(i\\sin(\\theta))^k<\/pre><\/div>\n\n\n\n
<\/div>\n\n\n\n

Breaking apart the sum into real and imaginary terms by moving all the even terms to one sum and odd terms to another we get:<\/p>\n\n\n\n

<\/div>\n\n\n\n
=\\sum_{k=0}^{n\/2}\\binom{n}{2k}\\cos^{n-2k}(\\theta)(i\\sin(\\theta))^{2k} +\\sum_{k=0}^{(n-1)\/2}\\binom{n}{2k+1}\\cos^{n-2k-1}(\\theta)(i\\sin(\\theta))^{2k+1}<\/pre><\/div>\n\n\n\n
<\/div>\n\n\n\n

Finally, we can pull i out of the right-side sum and write \\( i^{2k}=(\\sqrt{-1})^k \\) to get our final formula:<\/p>\n\n\n\n

=\\sum_{k=0}^{n\/2}(-1)^k\\binom{n}{2k}\\cos^{n-2k}(\\theta)\\sin(\\theta)^{2k} +i\\sum_{k=0}^{(n-1)\/2}(-1)^k\\binom{n}{2k+1}\\cos^{n-2k-1}(\\theta)\\sin(\\theta)^{2k+1} <\/pre><\/div>\n\n\n\n

We can test this for a couple of values of n and prove that they work: For example, n = 2 yields the familiar double-angle formulas for sine and cosine:<\/p>\n\n\n\n

\\sum_{k=0}^{2}(-1)^k\\binom{2}{2k}\\cos^{2-2k}(\\theta)\\sin(\\theta)^{2k} +i\\sum_{k=0}^{2}(-1)^k\\binom{2}{2k+1}\\cos^{2-2k-1}(\\theta)\\sin(\\theta)^{2k+1}<\/pre><\/div>\n\n\n\n

Hope this was helpful!<\/p>\n\n\n\n

<\/div>\n\n\n\n
(x+y)^n=\\binom{n}{0}x^n+\\binom{n}{1}x^{n-1}y+\\binom{n}{2}x^{n-2}y^2+\\ldots=\\sum_{k=0}^{n}\\binom{n}{k}x^{n-k}y^k<\/pre><\/div>\n\n\n\n
\"\"<\/figure>\n\n\n\n

The Binomial Theorem with Pascal’s Triangle.
Credit: [Mathematics Libretexts]<\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

For those of you who tried to derive the binomial expansion formula yourselves, here’s the step-by-step process below, based on how I derived the result when first studying complex numbers last year: Setting we can expand below: Breaking apart the sum into real and imaginary terms by moving all the even terms to one sum […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"nf_dc_page":"","_gspb_post_css":"","inline_featured_image":false,"_uag_custom_page_level_css":"","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":"","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":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","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":""},"class_list":["post-1078","page","type-page","status-publish","hentry"],"yoast_head":"\nTangent: Multiple Angle Formula for Sine\/Cosine - 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