sin s are rational, so..cos s are rational
$sin\ A,sin\ B,sin\ C$ of a triangle are rational. Prove that $cos\ A,cos\ B,cos\ C$ are also rational.
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Re: sin s are rational, so..cos s are rational
let the opposite sides of $A,B,C$ are $a,b,c$
through cos law
\[cos A=\frac{a^2-b^2-c^2}{-2bc}....(1)\]
if $a,b,c$ are rational,(by 1)then cos of all angles will be rational.
so we need to prove when $a,b,c$ are irrational.
through sine law\[\frac{sinA}{sinB}=\frac{a}{b},\frac{sinA}{sinC}=\frac{a}{c},\frac{sinB}{sinC}=\frac{b}{c}\]
that implies ratio among the sides are rational.so they have a common irrational divisor.
let $a=ma',b=mb',c=mc'$ where m is the common irrational divisor and $a',b',c'$ are rational.
in (1)
\[cos A=\frac{m^2(a'^2-b'^2-c'^2))}{-2mb'.mc'}=\frac{a'^2-b'^2-c'^2}{-2b'c'}\]
and cos A is rational.
so cos of all angles are rational.
through cos law
\[cos A=\frac{a^2-b^2-c^2}{-2bc}....(1)\]
if $a,b,c$ are rational,(by 1)then cos of all angles will be rational.
so we need to prove when $a,b,c$ are irrational.
through sine law\[\frac{sinA}{sinB}=\frac{a}{b},\frac{sinA}{sinC}=\frac{a}{c},\frac{sinB}{sinC}=\frac{b}{c}\]
that implies ratio among the sides are rational.so they have a common irrational divisor.
let $a=ma',b=mb',c=mc'$ where m is the common irrational divisor and $a',b',c'$ are rational.
in (1)
\[cos A=\frac{m^2(a'^2-b'^2-c'^2))}{-2mb'.mc'}=\frac{a'^2-b'^2-c'^2}{-2b'c'}\]
and cos A is rational.
so cos of all angles are rational.
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