Hey guys, i am doing a problem and one of the terms is "sec(2pi + x)", so I made that "1/cos(2pi + x)", but i'm stuck. I don't know what cos(2pi + x) is equal to.

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- Dec 1st 2007, 11:28 AMCenturionMonkeyTrig identities
Hey guys, i am doing a problem and one of the terms is "sec(2pi + x)", so I made that "1/cos(2pi + x)", but i'm stuck. I don't know what cos(2pi + x) is equal to.

- Dec 1st 2007, 12:40 PMgalactus
$\displaystyle sec(2{\pi}+x)=sec(x)$

- Dec 1st 2007, 12:51 PMCenturionMonkey
thanks. so does that also mean:

sin(2pi + x) = sin(x)

cos(2pi + x) = cos(x)

tan(2pi + x) = tan(x)

csc(2pi + x) = csc(x)

cot(2pi + x) = cot(x)

?

we weren't taught the (2x + x) identities, only the (2x - 1) - Dec 1st 2007, 12:53 PMgalactusQuote:

thanks. so does that also mean:

sin(2pi + x) = sin(x)

cos(2pi + x) = cos(x)

tan(2pi + x) = tan(x)

csc(2pi + x) = csc(x)

cot(2pi + x) = cot(x)

- Dec 1st 2007, 06:38 PMCenturionMonkey
ok another one.

i've gotten the left side down to

cos(x/2) - sin(x/2)

_________________

cos(x/2) + sin(x/2)

and the final right side equals:

1 - tan(x/2)

__________

1 + tan(x/2)

please help...

ps. i'm sorry for making it look so noobish. I am not sure how you guys do that fancy way of writing the equation down. The way Galactus over there answered me. - Dec 1st 2007, 06:57 PMJhevon
- Dec 1st 2007, 06:59 PMCenturionMonkey
yeah i would, but the left side i gave there, is like 8 steps down from the original left side. and it's too difficult to to go from right side, all the way to the original left side. :/

original LS: sec(x) - tan(x)

orignial RS: tan((pi/4) - (x/2)) - Dec 1st 2007, 07:05 PMJhevon
- Dec 1st 2007, 07:08 PMCenturionMonkey
oh i see what you're saying. alright, thanks a lot.