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)))sin(π12)=(√3−12√2)
cos((11pi)/12)=-((1+sqrt(3))/(2sqrt(2)))cos(11π12)=−(1+√32√2)
tan ((7pi)/12)=-2-sqrt(3)tan(7π12)=−2−√3

Explanation:

sin (pi/12)sin(π12)
=sin(pi/3 - pi/4)=sin(π3−π4)
=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)=sin(π3)cos(π4)−cos(π3)sin(π4)
=(sqrt(3)/2)(1/sqrt(2))-(1/2)(1/sqrt(2))=(√32)(1√2)−(12)(1√2)
=((sqrt(3)-1)/(2sqrt(2)))=(√3−12√2)

cos((11pi)/12)cos(11π12)
=cos((2pi)/3+pi/4)=cos(2π3+π4)
=cos((2pi)/3)cos(pi/4)-sin((2pi)/3)sin(pi/4)=cos(2π3)cos(π4)−sin(2π3)sin(π4)
=cos(pi-pi/3)cos(pi/4)-sin(pi-pi/3)sin(pi/4)=cos(π−π3)cos(π4)−sin(π−π3)sin(π4)
=-cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)=−cos(π3)cos(π4)−sin(π3)sin(π4)
=-(1/2)(1/sqrt(2))-(sqrt(3)/2)(1/sqrt(2))=−(12)(1√2)−(√32)(1√2)
=-((1+sqrt(3))/(2sqrt(2)))=−(1+√32√2)

tan ((7pi)/12)tan(7π12)
=tan(pi/3 + pi/4)=tan(π3+π4)
=(sqrt(3) + 1)/(1 - sqrt(3))=√3+11−√3
=((sqrt(3) + 1)(1+sqrt(3)))/(1 - 3)=(√3+1)(1+√3)1−3
=-2-sqrt(3)=−2−√3

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