{"id":7803,"date":"2022-09-20T20:32:32","date_gmt":"2022-09-20T20:32:32","guid":{"rendered":"https:\/\/mdr.foobrdigital.com\/?p=7803"},"modified":"2022-09-20T20:32:32","modified_gmt":"2022-09-20T20:32:32","slug":"half-angle-formulas","status":"publish","type":"post","link":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/2022\/09\/20\/half-angle-formulas\/","title":{"rendered":"Half Angle Formulas?"},"content":{"rendered":"\n<p>In this section, we will see the half angle formulas of sin, cos, and tan. We know the values of the trigonometric functions (sin, cos , tan, cot, sec, cosec) for the angles like 0\u00b0, 30\u00b0, 45\u00b0, 60\u00b0, and 90\u00b0 from the\u00a0trigonometric table. But to know the exact values of sin 22.5\u00b0, tan 15\u00b0, etc, the half angle formulas are extremely useful. Also, they are helpful in proving several\u00a0trigonometric identities. We have half angle formulas that are derived from the\u00a0double angle formulas\u00a0and they are\u00a0expressed in terms of half angles like \u03b8\/2, x\/2, A\/2, etc. Here is the list of important half angle formulas:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/mdr.foobrdigital.com\/wp-content\/uploads\/2022\/09\/uawaaaa.png\" alt=\"\" class=\"wp-image-7806\"\/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Half Angle Identities<\/h2>\n\n\n\n<p>Here are the popular&nbsp;<strong>half angle identities<\/strong>&nbsp;that we use in solving many trigonometry problems are as follows:<\/p>\n\n\n\n<ul><li><strong>Half angle formula of sin:&nbsp;<\/strong>sin A\/2 = \u00b1\u221a[(1 &#8211; cos A) \/ 2]<\/li><li><strong>Half angle formula of cos:&nbsp;<\/strong>cos A\/2 = \u00b1\u221a[(1 + cos A) \/ 2]<\/li><li><strong>Half angle formula of tan:&nbsp;<\/strong>tan A\/2 = \u00b1\u221a[1 &#8211; cos A] \/ [1 + cos A] (or) sin A \/ (1 + cos A) (or) (1 &#8211; cos A) \/ sin A<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Half Angle Formulas Derivation Using Double Angle Formulas<\/h2>\n\n\n\n<p>To derive the above formulas, first, let us derive the following half angle formulas. The\u00a0double angle formulas\u00a0are in terms of the double angles like 2\u03b8, 2A, 2x, etc. We know that the double angle formulas of\u00a0sin,\u00a0cos, and\u00a0tan\u00a0are<\/p>\n\n\n\n<ul><li>sin 2x\u00a0= 2 sin x cos x<\/li><li>cos 2x = cos<sup>2<\/sup>\u00a0x &#8211; sin<sup>2<\/sup>\u00a0x (or)<br>= 1 &#8211; 2 sin<sup>2<\/sup>x (or)<br>= 2 cos<sup>2<\/sup>x &#8211; 1<\/li><li>tan 2x = 2 tan x \/ (1 &#8211; tan<sup>2<\/sup>x)<\/li><\/ul>\n\n\n\n<p>If we replace x with A\/2 on both sides of every equation of double angle formulas, we get half angle identities (as 2x = 2(A\/2) = A).<\/p>\n\n\n\n<ul><li>sin A = 2 sin(A\/2) cos(A\/2)<\/li><li>cos A = cos<sup>2<\/sup>&nbsp;(A\/2) &#8211; sin<sup>2<\/sup>&nbsp;(A\/2) (or)<br>= 1 &#8211; 2 sin<sup>2<\/sup>&nbsp;(A\/2) (or)<br>= 2 cos<sup>2<\/sup>(A\/2) &#8211; 1<\/li><li>tan A = 2 tan (A\/2) \/ (1 &#8211; tan<sup>2<\/sup>(A\/2))<\/li><\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/mdr.foobrdigital.com\/wp-content\/uploads\/2022\/09\/half-angle-formula-derivation-dddddddddddddddddd-1.png\" alt=\"\" class=\"wp-image-7805\"\/><\/figure>\n\n\n\n<p>We can also derive one half angle formula using another half angle formula. For example, just from the formula of cos A, we can derive 3 important half angle identities for sin, cos, and tan which are mentioned in the first section. Here is the half angle formulas proof.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Half Angle Formula of Sin Proof<\/h2>\n\n\n\n<p>Now, we will prove the half angle formula for the sine function. Using one of the above formulas of cos A, we have<\/p>\n\n\n\n<p>cos A = 1 &#8211; 2 sin<sup>2<\/sup>&nbsp;(A\/2)<\/p>\n\n\n\n<p>From this,<\/p>\n\n\n\n<p>2 sin<sup>2<\/sup>&nbsp;(A\/2) = 1 &#8211; cos A<\/p>\n\n\n\n<p>sin<sup>2<\/sup>&nbsp;(A\/2) = (1 &#8211; cos A) \/ 2<\/p>\n\n\n\n<p>sin (A\/2) = \u00b1\u221a[(1 &#8211; cos A) \/ 2]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Half Angle Formula of Cos Derivation<\/h2>\n\n\n\n<p>Now, we will prove the half angle formula for the cosine function. Using one of the above formulas of cos A,<\/p>\n\n\n\n<p>cos A = 2 cos<sup>2<\/sup>(A\/2) &#8211; 1<\/p>\n\n\n\n<p>From this,<\/p>\n\n\n\n<p>2 cos<sup>2<\/sup>(A\/2) = 1 + cos A<\/p>\n\n\n\n<p>cos<sup>2<\/sup>&nbsp;(A\/2) = (1 + cos A) \/ 2<\/p>\n\n\n\n<p>cos (A\/2) = \u00b1\u221a[(1 + cos A) \/ 2]<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Half Angle Formula of Tan Derivation<\/h2>\n\n\n\n<p>We know that tan (A\/2) = [sin (A\/2)] \/ [cos (A\/2)]<\/p>\n\n\n\n<p>From the half angle formulas of sin and cos,<\/p>\n\n\n\n<p>tan (A\/2) = [\u00b1\u221a(1 &#8211; cos A)\/2] \/ [\u00b1\u221a(1 + cos A)\/2]<\/p>\n\n\n\n<p>=&nbsp;<strong>\u00b1\u221a[(1 &#8211; cos A)&nbsp;\/ (1 + cos A)]<\/strong><\/p>\n\n\n\n<p>This is one of the formulas of tan (A\/2). Let us derive the other two formulas by\u00a0rationalizing the denominator\u00a0here.<\/p>\n\n\n\n<p>tan (A\/2) = \u00b1\u221a[(1 &#8211; cos A) \/ (1 + cos A)] \u00d7 \u221a[(1 &#8211; cos A) \/ (1 &#8211; cos A)]<\/p>\n\n\n\n<p>= \u221a[(1 &#8211; cos A)<sup>2<\/sup>&nbsp;\/ (1 &#8211; cos<sup>2<\/sup>A)]<\/p>\n\n\n\n<p>= \u221a[(1 &#8211; cos A)<sup>2<\/sup>\/ sin<sup>2<\/sup>A]<\/p>\n\n\n\n<p>=&nbsp;<strong>(1 &#8211; cos A) \/ sin A<\/strong><\/p>\n\n\n\n<p>This is the second formula of tan (A\/2). To derive another formula, let us multiply and divide the above formula by (1 + cos A). Then we get<\/p>\n\n\n\n<p>tan (A\/2) = [(1 &#8211; cos A) \/ sin A] \u00d7 [(1 + cos A) \/ (1 + cos A)]<\/p>\n\n\n\n<p>= (1 &#8211; cos<sup>2<\/sup>A) \/ [sin A (1 + cos A)]<\/p>\n\n\n\n<p>= sin<sup>2<\/sup>A \/ [sin A (1 + cos A)]<\/p>\n\n\n\n<p>=&nbsp;<strong>sin A \/ (1 + cos A)<\/strong><\/p>\n\n\n\n<p>Thus, tan (A\/2) = \u00b1\u221a[(1 &#8211; cos A) \/ (1 + cos A)] = (1 &#8211; cos A) \/ sin A = sin A \/ (1 + cos A).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this section, we will see the half angle formulas of sin, cos, and tan. We know the values of the trigonometric functions (sin, cos , tan, cot, sec, cosec) for the angles like 0\u00b0, 30\u00b0, 45\u00b0, 60\u00b0, and 90\u00b0 from the\u00a0trigonometric table. But to know the exact values of sin 22.5\u00b0, tan 15\u00b0, etc, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[229],"tags":[],"_links":{"self":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7803"}],"collection":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/comments?post=7803"}],"version-history":[{"count":0,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7803\/revisions"}],"wp:attachment":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/media?parent=7803"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/categories?post=7803"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/tags?post=7803"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}