How do you find exact value of #csc (-7pi/ 3)#?

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Answer 1 · 2016-09-22T14:54:20

#csc(-(7pi)/3)=-2/sqrt3=-1.155#

All trigonometric functions have a cycle of #2pi#

Hence #csc(-(7pi)/3)#

= #csc(-(7pi)/3+2pi)#

= #csc(-(7pi)/3+(6pi)/3)#

= #csc(-pi/3)=-csc(pi/3)#

= #-2/sqrt3#

= #-(2xxsqrt3)/(sqrt3xxsqrt3)#

= #(-2xx1.732)/3=-3.464/3=-1.155#