How do you give the six trigonometric function values of (-3pi)/2?

1 Answer
Shwetank Mauria
Dec 11, 2016

#sin((-3pi)/2)=1#, #cos((-3pi)/2)=0#, #tan((-3pi)/2)=oo#, #cot((-3pi)/2)=0#, #sec((-3pi)/2)=oo# and #csc((-3pi)/2)=1#

Explanation:

All trigometrical ratios have a cycle of #2pi# i.e. their values repeat after every #2pi#.

Hence trigometrical ratios of #(-3pi)/2# and #2pi+(-(3pi)/2)# i.e. #(4pi-3pi)/2=pi/2# will be same.

As #sin(pi/2)=1#, #cos(pi/2)=0#, #tan(pi/2)=oo#, #cot(pi/2)=0#, #sec(pi/2)=oo# and #csc(pi/2)=1#

We have #sin((-3pi)/2)=1#, #cos((-3pi)/2)=0#, #tan((-3pi)/2)=oo#, #cot((-3pi)/2)=0#, #sec((-3pi)/2)=oo# and #csc((-3pi)/2)=1#

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