{"id":310,"date":"2026-04-18T07:29:57","date_gmt":"2026-04-18T07:29:57","guid":{"rendered":"https:\/\/qpr.ca\/blogs\/mathstuff\/?p=310"},"modified":"2026-04-18T17:04:52","modified_gmt":"2026-04-18T17:04:52","slug":"finding-cospi-15","status":"publish","type":"post","link":"https:\/\/qpr.ca\/blogs\/mathstuff\/2026\/04\/18\/finding-cospi-15\/","title":{"rendered":"Finding $#\\cos{\\pi\/15}#$"},"content":{"rendered":"\n

A recent Quora question asks: How can you prove that [math]\\cos{(\\pi\/15)} = (-1+\\sqrt{5}+\\sqrt{30+6\\sqrt{5}})\/4[\/math]? <\/a><\/em><\/p>\n\n\n\n

Well clearly you cannot<\/strong> prove that, because the RHS is bigger than sqrt(30)\/4 which is bigger than sqrt(16)\/4 which is 1.<\/p>\n\n\n\n

But to evaluate $#\\cos(\\pi\/15)#$ in terms of radicals a natural first step is to deal with the factor of 5 in the denominator by noting that $#\\cos(5\\theta)=\\cos(4\\theta +\\theta)#$ and combine the cosine sum rule with two applications of the double angle rule to get the well known result that<\/p>\n\n\n\n

$#\\begin{align}\\cos(5\\theta)&=(((\\cos^2(\\theta)-\\sin^2(\\theta))^2-(2\\sin(\\theta)\\cos(\\theta))^2)\\cos(\\theta)\\\\&-2(2\\sin(\\theta)\\cos(\\theta))(\\cos^2(\\theta)-\\sin^2(\\theta))\\sin(\\theta)\\\\&=((2(2\\cos^2(\\theta)-1)^2\u20131)\\cos(\\theta)-2(2\\sin(\\theta)\\cos(\\theta))(2\\cos^2(\\theta)-1)\\sin(\\theta)\\\\&=((2(2\\cos^2(\\theta)-1)^2\u20131)\\cos(\\theta)-4(1-\\cos^2(\\theta)\\cos(\\theta))(2\\cos^2(\\theta)-1)\\\\&=16\\cos^5(\\theta)-20\\cos^3(\\theta)+5\\cos(\\theta)\\end{align}#$<\/p>\n\n\n\n

Now one approach would be to use this with $#5\\theta=\\pi\/3#$ to get the equation $#16\\cos^5(\\pi\/15)-20\\cos^3(\\pi\/15)+5\\cos(\\pi\/15)=cos(\\pi\/3)=1\/2#$<\/p>\n\n\n\n

But the 5th degree equation may be easier if we look first at $#5\\theta=2\\pi#$ to get<\/p>\n\n\n\n

$#16\\cos^5(2\\pi\/5)-20\\cos^3(2\\pi\/5)+5\\cos(2\\pi\/5)=\\cos(2\\pi)=1#$<\/p>\n\n\n\n

For $#c=\\cos(2\\pi\/5)#$ this gives $#16c^5\u201320c^3+5c-1=0#$ for which $#c=1#$ is clearly a solution so we can reduce the degree by long division to get $#16c^4+16c^3\u20134c^2\u20134c+1=0#$<\/p>\n\n\n\n

Since this is of just 4th degree we know that there is an explicit solution in radicals which we can write down using Cardano\u2019s formula; but I just noticed that the two highest degree terms match those of $#(4c^2+2c)^2=16c^4+16c^3+4c^2#$, and if we look at what\u2019s left we see something interesting – namely $#16c^4+16c^3\u20134c^2\u20134c+1=(4c^2+2c)^2-8c^2-4c+1=(4c^2+2c)^2-2(4c^2+2c)+1=(4c^2+2c-1)^2#$.<\/p>\n\n\n\n

So the only possibilities are for $#c#$ to be one of the roots of $#4c^2+2c-1#$ which are $#\\frac{-1\\pm \\sqrt{5}}{4}#$ and only the $#+#$ sign gives a possible value for $#\\cos#$ so we must have $#\\cos(2\\pi\/5)=\\frac{ \\sqrt{5}-1}{4}#$.<\/p>\n\n\n\n

And since $#\\pi\/15=2\\pi\/5-\\pi\/3#$ we can now easily use the cosine of a difference rule to get<\/p>\n\n\n\n

$#\\cos(\\pi\/15)=(-1+\\sqrt{5}+\\sqrt{30+6\\sqrt{5}})\/8#$.<\/p>\n\n\n\n

So the suggested answer was just off by a factor of 2.<\/p>\n\n\n\n

Source: (1002) Alan Cooper’s answer to How can you prove that [math]cos{(pi\/15)} = (-1+sqrt{5}+sqrt{30+6sqrt{5}})\/4[\/math]? – Quora<\/a><\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

A recent Quora question asks: How can you prove that [math]\\cos{(\\pi\/15)} = (-1+\\sqrt{5}+\\sqrt{30+6\\sqrt{5}})\/4[\/math]? Well clearly you cannot prove that, because the RHS is bigger than sqrt(30)\/4 which is bigger than sqrt(16)\/4 which is 1. But to evaluate $#\\cos(\\pi\/15)#$ in terms of radicals a natural first step is to deal with the factor of 5 in…<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-310","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/posts\/310","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/comments?post=310"}],"version-history":[{"count":8,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/posts\/310\/revisions"}],"predecessor-version":[{"id":338,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/posts\/310\/revisions\/338"}],"wp:attachment":[{"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/media?parent=310"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/categories?post=310"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/mathstuff\/wp-json\/wp\/v2\/tags?post=310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}