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add them?

so what do i add?

That's where i'm cofused

¬_¬
Sin 150 = Sin 30
Sin 135 = Sin 45

I do not get it.

Do you not get why that is true?

Or do you not get that sin(150 + 135) = sin 285

both

ok thanks

-90?

Last line is -sin(75).
Where did you get 90 from?

i have no idea what -sin 35= ?

the answer is?

any *

Just admit you were wrong, please, and I won't have to own you.

lol, own who?

the graph is actually easier, never realized it :p

LOL, do you know how yo use it loco :)?

lol, yes it will :) , you've never used it right?

kind of, ta showed it to us. still iffy on it but practicing.

besides loco, the information INew has given you also is valuable, work with both ^_^

thanks guys :)

np ^_^

but did you figure it out ? :)

stop fighting guys

Simplify the trigonometric expression.
cos[x]/sec[x] + sin[x]/csc[x]

Sorry, I thought they were implying they could do the hard part in a circle, not the easy part :/

I'm not fighting LOL, we're discussing loco ^_^

1/sec = cos. 1/csc = sin. cos * 1/sec = .... etc

how about we i mean u guys solve this problem

you still didn't solve it? .-.

no, i hate this stuff

don't say that, you still didn't master it , but you will with practice and it'll be easy on you ^_^

i've been practicing a lot, but still having trouble :'(

\[\frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2x\]

it's alright, keep on practicing and you'll master it :)

sstarica help?

hold on :)

newton it's wrong

What's wrong? (Before answering, note that I am never wrong)

sorry, i thought that was the answer

try to explain it in a simple way :)

He meant I am wrong about the second problem he posted in here. I only did half of it

Yes, my FIRST method was comlpicated, but my second was amazing.

he posted another question?

lol, good that you've made it easier ^_^

Thankfully still no graphs required

oh that one , LOL LOCO! you've posted it in a different thread =D and I was writing it down

u can write it here

nvm it, INew answered it though :)

it's not right though

you are right :) i got the same answer

No, you aren't wrong. Unless you count not using \frac{}{} for fractions as wrong (I personally do)

well done ^_^

lol, I didn't get you INew

\[\frac{1}{2} \geq 1/2\]

hmm, I can't find the frac() symbol lol >_<

still don't know how to make that fraction line ^_^"

last problem....
Simplify the trigonometric expression.
cos[x]/ tan[x] + sex[x]

frac{a}{b} gives\[\frac{a}{b}\]

oh, thank you :)

sex(x) - lolololololol

LOL

sec[x], sorry :P

wish math was that fun

I general, always write tan as sin/cos , sec as 1/cos etc.

When trying to simplify things.

ORLY?

maybe? :) correct me lol

it's not correct

\[\cos^2x + 1 \not= \sin^2x\]

hmm, where is my mistake

LOL! right >_< sorry

hmm, solve it :)

man, u guys are at it.

LOL, loco we're not fighting =P, hold on :)

Ugh, it doesn't simplify.. at all. But also sec x = 1/cos(x) ... >_> sorry

\[\sec x + \csc x -\sin x\]
.. but I'm tired.

Loco left is for a new thread. Un coooooooooooool

i posted it in a new thread also, but u guys were to busy

2nd one

Was my answer right?

hold on

Joke, I know it was. Just unsure what the 'expected' form is.

it's wrong newton

I'm not sure though, I was laggin ^_^|"

just go over it please :)

nope

wrong? ._.

yep

blah~, I'll try again ^_^

sec(x) + cos(x)cot(x), maybe?

don't know, my school's hw program sucks nuts

i only get 3 tries per problem

at least it's not timed, i've been doing this since 9pm

it really pisses me off and the professor is a b**** and doesn't do anything about it

lol, both of you calm down and just try again ^_^

\[\frac{\cos x}{\tan x} + \sec x = \frac{\cos^2x}{\sin x} + \sec x = \cos x\cot x + \sec x = ?\]

Which can also be written as sec x + csc x - sin x , and many other things.

i'll wait for sstarica's answer

True, she (I think, but maybe he?) does have more fans than me, so that's a good call.

I'm a she =P

I'm a she =P

r u hot? ;)

just to make sure, what is the question loco? :)

let me find it

Simplify the trigonometric expression.
cos[x]/ tan[x] + sec[x]

what on earth? lol what kind of quetion is that? >_>

idk.... curious loco mind

Don't worry about loco the sexpest . But in answer to the question of course she is, she does Maths.

ok... lets continue with math. i don't wanna go to jail

I got :
cscx -sinx + secx *sigh* not sure though, but that's what I got

cos^2x + 1
----------
sinx

Blexting is about to say : This problem is trivial via the use of synthetic division.

Blexting sec x is not 1 / sin x, btw.

it's wrong blexting

lol

check mine please ^^" if I'm wrong, then I'm sorry

My answer (which sstarica now has, too) has been right all along.

ok this the last try i have left any final answers

lol INew, you got the same answer? :)

yeah he did or she

Oh, I'm a she. Hence the Newton name and everythi-- oh, wait a second ...

good question ._.

nope it didn't work. sorry guys. told u the system is bs

=-=......I'm sorry

i got a 90 on the quiz though

excellent :)

Out of interest, what level is the quiz? As in, what educational level?

THANKS FOR TRYING NEWTON (HE OR SHE?) AND SSTARICA (SEND ME UR # ;p)

PRE-CALC

lok, someone answered this question in the other thread >_< LOL!

but the professor is really tough, doesn't curve or anything

american ed system succks retricemate

Don't worry, the person on the other thread got it wrong.

bloody teachers

yeah he did

THANKS SSTARICA :)

for my own personal interest, where have you reaced?

np :)

what do u mean

i think it can mean both. depends on what you're talking about