Question 975681


Use the formulas for trigonometric functions of half argument &nbsp;(see, &nbsp;for example, &nbsp;the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Compendium-of-Trigonometry-Formulas.lesson>FORMULAS FOR TRIGONOMETRIC FUNCTIONS</A>&nbsp; in this site)


{{{sin(beta/2)}}} = {{{sqrt((1-cos^2(beta))/2)}}} = {{{sqrt((1-((-2)/7)^2)/2)}}} = {{{sqrt((1-(4/49))/2)}}} = {{{sqrt(((49-4)/49)/2)}}} = {{{sqrt(45/(49*2))}}} = {{{(3/7)*sqrt(5/2)}}} = {{{(3/7)*(1/2)*sqrt(10)}}} = {{{(3/14)*sqrt(10)}}}. 


{{{cos(beta/2)}}} = {{{-sqrt((1+cos^2(beta))/2)}}} = {{{-sqrt((1+((-2)/7)^2)/2)}}} = {{{-sqrt((1+(4/49))/2)}}} = {{{-sqrt(((49+4)/49)/2)}}} = {{{-sqrt(53/(49*2))}}} = {{{-(1/7)*sqrt(53/2)}}} = {{{-(1/7)*(sqrt(106)/2)}}} = {{{-(1/14)*sqrt(106)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(the sign &nbsp;"minus" &nbsp;because the angle &nbsp;{{{beta/2}}}&nbsp; is in the 2-nd quadrant). 


{{{tan(beta/2)}}} = {{{sin(beta/2)/cos(beta/2)}}} = {{{-((3*14)/14)*(sqrt(10)/sqrt(106))}}} = {{{-3*(sqrt(10)/sqrt(106))}}}.