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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