How do you find the 6 trigonometric functions for -240 degrees?

1 Answer
Jul 1, 2018

Please see below.

Explanation:

Here, #-240^circ=(-180^circ)+(-60^circ)=>II^(nd)Quadrant#

But , #+240^circ=180^circ+60^circ=>III^(rd)Quadrant#

So ,

#color(red)(sinx,cosx ,secx,cscx ,are-ve )andcolor(blue)( tanx,cotx,are +ve#

#(1)sin(-240^circ)=-sin240^circto[becausesin(-theta)=-sintheta]#
#color(white)(=>sin(-240^circ))=-sin(180^circ+60^circ)toIII^(rd)Quadrant#
#color(white)(=>sin(-240^circ))=-(-sin60^circ)#
#=>sin(-240^circ)=sin60^circ=sqrt3/2#

#(2)cos(-240^circ)=cos240^circto[becausecos(-theta)=costheta]#
#color(white)(=>sin(-240^circ))=cos(180^circ+60^circ)toIII^(rd)Quadrant#
#=>cos(-240^circ)=-cos60^circ=-1/2#

#(3)csc(-240^circ)=-csc240^circto[becausecsc(-theta)=-csctheta]#
#color(white)(=>sin(-240^circ))=-csc(180^circ+60^circ)toIII^(rd)Quadrant#
#color(white)(=>sin(-240^circ))=-(-csc60^circ)#
#=>csc(-240^circ)=csc60^circ=2/sqrt3#

#(4)sec(-240^circ)=sec240^circto[becausesec(-theta)=sectheta]#
#color(white)(=>sin(-240^circ))=sec(180^circ+60^circ)toIII^(rd)Quadrant#
#=>sec(-240^circ)=-sec60^circ=-2#

#(5)tan(-240^circ)=-tan240^circto[becausetan(-theta)=-tantheta]#
#color(white)(=>sin(-240^circ))=-tan(180^circ+60^circ)toIII^(rd)Quadrant#
#color(white)(=>sin(-240^circ))=-tan60^circ#
#=>tan(-240^circ)=-tan60^circ=-sqrt3#

#(6)cot(-240^circ)=-cot240^circto[becausecot(-theta)=-cottheta]#
#color(white)(=>sin(-240^circ))=-cot(180^circ+60^circ)toIII^(rd)Quadrant#
#color(white)(=>sin(-240^circ))=-cot60^circ#
#=>cot(-240^circ)=-cot60^circ=-1/sqrt3#