If #cos (pi/6) = (sqrt 3/2)#, how can I find #cos (pi/12)# using a double angle formula?

1 Answer
Dec 2, 2017

#cos(pi/12)=1/2sqrt(sqrt3+2)#

Explanation:

#"using the "color(blue)"double angle formula"#

#•color(white)(x)cos2A=2cos^2A-1#

#cos(2xxpi/12)=2cos^2(pi/12)-1#

#rArrcos(pi/6)=2cos^2(pi/12)-1#

#rArrcos^2(pi/12)=1/2(cos(pi/6)+1)#

#color(white)(rArrcos^2(pi/12))=1/2(sqrt3/2+1)#

#color(white)(rArrcos^2(pi/12))=1/4(sqrt3+2)#

#rArrcos(pi/12)=sqrt(1/4(sqrt3+2)#

#color(white)(rxxxxxxxx)=1/2sqrt(sqrt3+2)larrcolor(red)"exact value"#