How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle #pi/12#?

1 Answer
Jim G.
Aug 6, 2017

#1/2sqrt(2-sqrt3),1/2sqrt(2+sqrt3),sqrt(7-4sqrt3)#

Explanation:

#"since "pi/12" is in the first quadrant all of the trig. ratios"#
#"will be positive"#

#•color(white)(x)sin(x/2)=+-sqrt((1-cosx)/2)#

#"here "(x/2)=pi/12rArrx=pi/6#

#sin(pi/12)=+sqrt((1-cos(pi/6))/2#

#color(white)(xxxxxx)=sqrt((1-sqrt3/2)/2#

#color(white)(xxxxxx)=sqrt(((2-sqrt3))/4)=1/2sqrt(2-sqrt3)#
#color(blue)"---------------------------------------------------------"#

#•color(white)(x)cos(x/2)=+-sqrt((1+cosx)/2)#

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

#color(white)(xxxxxx)=sqrt((2+sqrt3)/4)=1/2sqrt(2+sqrt3)#
#color(blue)"----------------------------------------------------------"#

#•color(white)(x)tan(x/2)=+-sqrt((1-cosx)/(1+cosx))#

#tan(pi/12)=+sqrt((2-sqrt3)/(2+sqrt3))#

#"multiply the numerator/denominator by "(2-sqrt3)#

#rArrtan(pi/12)=sqrt(7-4sqrt3)#

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