Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

1 Answer
Joel Kindiak
Aug 19, 2015

#sin (pi/12)=((sqrt(3)-1)/(2sqrt(2)))#
#cos((11pi)/12)=-((1+sqrt(3))/(2sqrt(2)))#
#tan ((7pi)/12)=-2-sqrt(3)#

Explanation:

#sin (pi/12)#
#=sin(pi/3 - pi/4)#
#=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)#
#=(sqrt(3)/2)(1/sqrt(2))-(1/2)(1/sqrt(2))#
#=((sqrt(3)-1)/(2sqrt(2)))#

#cos((11pi)/12)#
#=cos((2pi)/3+pi/4)#
#=cos((2pi)/3)cos(pi/4)-sin((2pi)/3)sin(pi/4)#
#=cos(pi-pi/3)cos(pi/4)-sin(pi-pi/3)sin(pi/4)#
#=-cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)#
#=-(1/2)(1/sqrt(2))-(sqrt(3)/2)(1/sqrt(2))#
#=-((1+sqrt(3))/(2sqrt(2)))#

#tan ((7pi)/12)#
#=tan(pi/3 + pi/4)#
#=(sqrt(3) + 1)/(1 - sqrt(3))#
#=((sqrt(3) + 1)(1+sqrt(3)))/(1 - 3)#
#=-2-sqrt(3)#

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