Question 984661


There is the classic formula for the sines of the half argument:


{{{sin^2(alpha/2)}}} = {{{(1-cos(alpha))/2}}}. 


(see, &nbsp;for example, &nbsp;the lessons &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Compendium-of-Trigonometry-Formulas.lesson>FORMULAS FOR TRIGONOMETRIC FUNCTIONS</A>&nbsp; and &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A>&nbsp; in this site, &nbsp;or any systematic textbook on trigonometry).


From this formula 


{{{cos(alpha)}}} = {{{1 - 2sin^2(alpha/2)}}}. 


By applying this formula to our case, &nbsp;you will get


{{{1 - 2sin^2(3pi/5)}}} = {{{cos(6pi/5)}}} = {{{-cos(pi/5)}}} = -cos(36°).


<B>Answer</B>. &nbsp;{{{1 - 2sin^2(3pi/5)}}} = {{{-cos(pi/5)}}} = -cos(36°).