Verify the identity.<\/p>\n\n\n\n
sin2A cos2A = Sin2A – 4sin3A cosA<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To verify the identity:<\/p>\n\n\n\n [ We\u2019ll verify if the left-hand side (LHS)<\/strong> equals the right-hand side (RHS)<\/strong> using trigonometric identities.<\/p>\n\n\n\n [ Use the double-angle identities<\/strong>:<\/p>\n\n\n\n So, Now multiply: [ Use: So: Factor: We have:<\/p>\n\n\n\n We now show that: This is a known identity:<\/p>\n\n\n\n [ \u2705 Therefore: So both sides simplify to the same expression: [ This identity can be verified using standard trigonometric identities, particularly the double-angle formulas<\/strong> for sine and cosine. The double-angle identity for sine is (\\sin2A = 2\\sin A \\cos A), while for cosine there are multiple forms, including (\\cos2A = \\cos^2 A – \\sin^2 A) and (\\cos2A = 1 – 2\\sin^2 A). These identities are key to rewriting complex expressions involving angles like (2A) or higher powers of sine and cosine.<\/p>\n\n\n\n In this problem, we start by simplifying both sides independently. For the left-hand side (LHS), we expand (\\sin2A \\cos2A) by substituting the known identities. The resulting expression becomes (2\\sin A \\cos A (\\cos^2 A – \\sin^2 A)). Meanwhile, the right-hand side (RHS), (\\sin2A – 4\\sin^3 A \\cos A), can also be rewritten using the identity for (\\sin2A). After factoring, it becomes (2\\sin A \\cos A (1 – 2\\sin^2 A)).<\/p>\n\n\n\n To complete the verification, we show that the expressions inside the parentheses on both sides are equivalent. That is, we show that (\\cos^2 A – \\sin^2 A = 1 – 2\\sin^2 A), which is a known identity for (\\cos2A). Because both LHS and RHS simplify to the same exact expression, the identity is proven to be true.<\/p>\n","protected":false},"excerpt":{"rendered":" Verify the identity. sin2A cos2A = Sin2A – 4sin3A cosA The correct answer and explanation is : To verify the identity: [\\sin2A \\cos2A = \\sin2A – 4\\sin^3A \\cosA] We\u2019ll verify if the left-hand side (LHS) equals the right-hand side (RHS) using trigonometric identities. Step 1: Expand both sides using identities Left-hand side (LHS): [\\sin2A \\cos2A] […]<\/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-208006","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208006","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=208006"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208006\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208006"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208006"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208006"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\sin2A \\cos2A = \\sin2A – 4\\sin^3A \\cosA
]<\/p>\n\n\n\n
\n\n\n\nStep 1: Expand both sides using identities<\/h3>\n\n\n\n
Left-hand side (LHS):<\/h4>\n\n\n\n
\\sin2A \\cos2A
]<\/p>\n\n\n\n\n
[
\\sin2A \\cos2A = (2\\sin A \\cos A)(\\cos^2 A – \\sin^2 A)
]<\/p>\n\n\n\n
[
= 2\\sin A \\cos A (\\cos^2 A – \\sin^2 A)
]<\/p>\n\n\n\n
\n\n\n\nRight-hand side (RHS):<\/h4>\n\n\n\n
\\sin2A – 4\\sin^3 A \\cos A
]<\/p>\n\n\n\n
(\\sin2A = 2\\sin A \\cos A)<\/p>\n\n\n\n
[
\\sin2A – 4\\sin^3 A \\cos A = 2\\sin A \\cos A – 4\\sin^3 A \\cos A
]<\/p>\n\n\n\n
[
= 2\\sin A \\cos A (1 – 2\\sin^2 A)
]<\/p>\n\n\n\n
\n\n\n\nStep 2: Compare LHS and RHS<\/h3>\n\n\n\n
\n
[
\\cos^2 A – \\sin^2 A = 1 – 2\\sin^2 A
]<\/p>\n\n\n\n
\\cos 2A = \\cos^2 A – \\sin^2 A = 1 – 2\\sin^2 A
]<\/p>\n\n\n\n
[
\\cos^2 A – \\sin^2 A = 1 – 2\\sin^2 A
]<\/p>\n\n\n\n
[
2\\sin A \\cos A (1 – 2\\sin^2 A)
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/h3>\n\n\n\n
\\sin2A \\cos2A = \\sin2A – 4\\sin^3A \\cosA \\quad \\text{is a TRUE identity.}
]<\/p>\n\n\n\n
\n\n\n\n\ud83d\udd0d Summary Explanation (300+ words):<\/h3>\n\n\n\n