How would you find the exact value of the six trigonometric functions of 5 #pi#/6?

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Answer 1 · 2018-08-03T10:48:37

Please see below.

Here , #theta=(5pi)/6=pi-pi/6to2^(nd)Quadrant#

#color(red)(sintheta >0, csctheta > 0,costheta < 0,sectheta < 0,tantheta < 0,cottheta < 0#

#(1)sintheta=sin((5pi)/6)=sin(pi-pi/6)=sin(pi/6)=1/2#

#(2)costheta=cos((5pi)/6)=cos(pi-pi/6)=-cos(pi/6)=-sqrt3/2#

#(3)tantheta=tan((5pi)/6)=sin((5pi)/6)/cos((5pi)/6)=(1/2)/(-sqrt3/2)=-1/sqrt3#

#(4)csctheta=csc((5pi)/6)=1/sin((5pi)/6)=1/(1/2)=2#

#(5)sectheta=sec ((5pi)/6)=1/cos((5pi)/6)=1/(-sqrt3/2)=-2/sqrt3#

#(6)cottheta=cot((5pi)/6)=1/tan((5pi)/6)=1/(-1/sqrt3)=-sqrt3#