To find the exact value of sin\u2061(5\u03c08)\\sin\\left(\\frac{5\\pi}{8}\\right)sin(85\u03c0\u200b), we will use a combination of sum and difference identities and half-angle formulas.<\/p>\n\n\n\n
Step 1: Express 5\u03c08\\frac{5\\pi}{8}85\u03c0\u200b as a sum of known angles<\/h3>\n\n\n\n
We can write 5\u03c08\\frac{5\\pi}{8}85\u03c0\u200b as the sum of two known angles: 5\u03c08=\u03c02+\u03c08\\frac{5\\pi}{8} = \\frac{\\pi}{2} + \\frac{\\pi}{8}85\u03c0\u200b=2\u03c0\u200b+8\u03c0\u200b<\/p>\n\n\n\n
Step 2: Apply the sum formula for sine<\/h3>\n\n\n\n
We use the sum formula for sine: sin\u2061(A+B)=sin\u2061Acos\u2061B+cos\u2061Asin\u2061B\\sin(A + B) = \\sin A \\cos B + \\cos A \\sin Bsin(A+B)=sinAcosB+cosAsinB<\/p>\n\n\n\n
In this case, A=\u03c02A = \\frac{\\pi}{2}A=2\u03c0\u200b and B=\u03c08B = \\frac{\\pi}{8}B=8\u03c0\u200b. Substituting these values into the formula: sin\u2061(\u03c02+\u03c08)=sin\u2061(\u03c02)cos\u2061(\u03c08)+cos\u2061(\u03c02)sin\u2061(\u03c08)\\sin\\left(\\frac{\\pi}{2} + \\frac{\\pi}{8}\\right) = \\sin\\left(\\frac{\\pi}{2}\\right)\\cos\\left(\\frac{\\pi}{8}\\right) + \\cos\\left(\\frac{\\pi}{2}\\right)\\sin\\left(\\frac{\\pi}{8}\\right)sin(2\u03c0\u200b+8\u03c0\u200b)=sin(2\u03c0\u200b)cos(8\u03c0\u200b)+cos(2\u03c0\u200b)sin(8\u03c0\u200b)<\/p>\n\n\n\n
Step 3: Simplify using known values for sin\u2061(\u03c02)\\sin\\left(\\frac{\\pi}{2}\\right)sin(2\u03c0\u200b) and cos\u2061(\u03c02)\\cos\\left(\\frac{\\pi}{2}\\right)cos(2\u03c0\u200b)<\/h3>\n\n\n\n
We know that: sin\u2061(\u03c02)=1andcos\u2061(\u03c02)=0\\sin\\left(\\frac{\\pi}{2}\\right) = 1 \\quad \\text{and} \\quad \\cos\\left(\\frac{\\pi}{2}\\right) = 0sin(2\u03c0\u200b)=1andcos(2\u03c0\u200b)=0<\/p>\n\n\n\n
So, the expression simplifies to: sin\u2061(\u03c02+\u03c08)=1\u22c5cos\u2061(\u03c08)+0\u22c5sin\u2061(\u03c08)\\sin\\left(\\frac{\\pi}{2} + \\frac{\\pi}{8}\\right) = 1 \\cdot \\cos\\left(\\frac{\\pi}{8}\\right) + 0 \\cdot \\sin\\left(\\frac{\\pi}{8}\\right)sin(2\u03c0\u200b+8\u03c0\u200b)=1\u22c5cos(8\u03c0\u200b)+0\u22c5sin(8\u03c0\u200b)<\/p>\n\n\n\n
Thus, we have: sin\u2061(5\u03c08)=cos\u2061(\u03c08)\\sin\\left(\\frac{5\\pi}{8}\\right) = \\cos\\left(\\frac{\\pi}{8}\\right)sin(85\u03c0\u200b)=cos(8\u03c0\u200b)<\/p>\n\n\n\n
Step 4: Use the half-angle formula to find cos\u2061(\u03c08)\\cos\\left(\\frac{\\pi}{8}\\right)cos(8\u03c0\u200b)<\/h3>\n\n\n\n
Now, we need to compute cos\u2061(\u03c08)\\cos\\left(\\frac{\\pi}{8}\\right)cos(8\u03c0\u200b). To do this, we use the half-angle formula for cosine: cos\u2061(\u03b82)=\u00b11+cos\u2061(\u03b8)2\\cos\\left(\\frac{\\theta}{2}\\right) = \\pm \\sqrt{\\frac{1 + \\cos(\\theta)}{2}}cos(2\u03b8\u200b)=\u00b121+cos(\u03b8)\u200b\u200b<\/p>\n\n\n\n
Let \u03b8=\u03c04\\theta = \\frac{\\pi}{4}\u03b8=4\u03c0\u200b, so that \u03c08=\u03c04\u00d712\\frac{\\pi}{8} = \\frac{\\pi}{4} \\times \\frac{1}{2}8\u03c0\u200b=4\u03c0\u200b\u00d721\u200b. Then: cos\u2061(\u03c08)=1+cos\u2061(\u03c04)2\\cos\\left(\\frac{\\pi}{8}\\right) = \\sqrt{\\frac{1 + \\cos\\left(\\frac{\\pi}{4}\\right)}{2}}cos(8\u03c0\u200b)=21+cos(4\u03c0\u200b)\u200b\u200b<\/p>\n\n\n\n
We know that: cos\u2061(\u03c04)=22\\cos\\left(\\frac{\\pi}{4}\\right) = \\frac{\\sqrt{2}}{2}cos(4\u03c0\u200b)=22\u200b\u200b<\/p>\n\n\n\n
Substitute this into the half-angle formula: cos\u2061(\u03c08)=1+222=22+222=2+24=2+22\\cos\\left(\\frac{\\pi}{8}\\right) = \\sqrt{\\frac{1 + \\frac{\\sqrt{2}}{2}}{2}} = \\sqrt{\\frac{\\frac{2}{2} + \\frac{\\sqrt{2}}{2}}{2}} = \\sqrt{\\frac{2 + \\sqrt{2}}{4}} = \\frac{\\sqrt{2 + \\sqrt{2}}}{2}cos(8\u03c0\u200b)=21+22\u200b\u200b\u200b\u200b=222\u200b+22\u200b\u200b\u200b\u200b=42+2\u200b\u200b\u200b=22+2\u200b\u200b\u200b<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
Thus, the exact value of sin\u2061(5\u03c08)\\sin\\left(\\frac{5\\pi}{8}\\right)sin(85\u03c0\u200b) is: sin\u2061(5\u03c08)=2+22\\sin\\left(\\frac{5\\pi}{8}\\right) = \\frac{\\sqrt{2 + \\sqrt{2}}}{2}sin(85\u03c0\u200b)=22+2\u200b\u200b\u200b<\/p>\n\n\n\n Find the exact value of sin(5pi\/8) by using the formulas (sum and difference, half angle, double angle The Correct Answer and Explanation is: To find the exact value of sin\u2061(5\u03c08)\\sin\\left(\\frac{5\\pi}{8}\\right)sin(85\u03c0\u200b), we will use a combination of sum and difference identities and half-angle formulas. Step 1: Express 5\u03c08\\frac{5\\pi}{8}85\u03c0\u200b as a sum of known angles We can […]<\/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-240276","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240276","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=240276"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240276\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240276"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240276"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240276"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-117.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"