# Trig identities

• Dec 1st 2007, 11:28 AM
CenturionMonkey
Trig 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 PM
galactus
$\displaystyle sec(2{\pi}+x)=sec(x)$
• Dec 1st 2007, 12:51 PM
CenturionMonkey
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 PM
galactus
Quote:

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)
Yep.
• Dec 1st 2007, 06:38 PM
CenturionMonkey
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)

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 PM
Jhevon
Quote:

Originally Posted by CenturionMonkey
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)

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.

$\displaystyle \frac {1 - \tan (x/2)}{1 + \tan (x/2)} = \frac {1 - \frac {\sin (x/2)}{\cos (x/2)}}{1 + \frac {\sin (x/2)}{\cos (x/2)}}$

now combine the fractions in the numerator and denominator and simplify
• Dec 1st 2007, 06:59 PM
CenturionMonkey
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 PM
Jhevon
Quote:

Originally Posted by CenturionMonkey
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))

there is no need to go to the original left side. you will prove they are equivalent if you can bring them both to the same thing. you take advantage of the law of transitivity here.
• Dec 1st 2007, 07:08 PM
CenturionMonkey
oh i see what you're saying. alright, thanks a lot.