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

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

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