How do you evaluate #sec((9pi)/4)#?

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Answer 1 · 2016-09-24T15:00:45

#sec((9pi)/4)=sqrt2#

All trigonometric ratios have a cycle of #2pi#, i.e.

their values repeat after every #2pi# and we can say for example

#sinx=sin(2pi+x)=sin(4pi+x)=sin(6pi+x)=....# and

#secx=sec(2pi+x)=sec(4pi+x)=sec(6pi+x)=....#

Now #sec((9pi)/4)=sec((8pi)/4+pi/4)=sec(2pi+pi/4)=sec(pi/4)#

But #sec(pi/4)=sqrt2#, hence

#sec((9pi)/4)=sqrt2#