Use an addition or subtraction formula to find the exact value of the expression. sin(285)

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Use an addition or subtraction formula to find the exact value of the expression. sin(285)

Mathematics
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Well, lets assume you know the 'obvious' values of sin - {0, 30, 45, 60, 90} + 360n. How can you use them to make 285?
add them?
Indeed. Use sin(a+b) with a couple of them and you should be able to work it out. I guess the main problem is how to do it most efficiently.

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so what do i add?
That's where i'm cofused
Something to consider is that sin (180 - x) = sin x (the best way is probably to add 2 values of sin between 100 and 180
¬_¬ 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
Well, if 0 < x < 180 , then sin (180 - x) = sin x Hence the above claim. And 135 + 150 = 285 (clearly) sin(150 + 135) = sin(150)cos(135) + sin(135)cos(150) = sin(30)cos(135) + sin(45)(cos(150) All you need to do is use a SIMILAR rule one the cos values and you're done. I'll leave that to you. NB this is how I'd do it, but there may be a quicker way.
ok thanks
Gah, actually I prefer: Sin 285 = sin (285 - 360) = sin(-75) Sin is an odd function => sin(-75) = -sin(75) sin 75 = sin(30+45) = blah. -sin 35 = -blah.
-90?
Last line is -sin(75). Where did you get 90 from?
i have no idea what -sin 35= ?
It's meant to be -sin(75) , sorry! But sin (75) = sin(45 + 30) = sin(45)cos(30) + sin(30)cos(45) And sin(285) = sin(-75) = -sin(75) = -(sin(45)cos(30) + sin(30)cos(45)) Because sin(-a) = -sin(a)
the answer is?
hmm, INew, wouldn't it be easier if you just drew the graph and compute it from the graph? :) it'll be alot easier ^_^
sin(45) = cos(45) = 1/sqrt(2) sin(30) = 1/2 cos(30) = sqrt(3)/2 \[\implies -\sin(75) = -\frac{1}{\sqrt{2}} \cdot \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = -\frac{1+ \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{2} + \sqrt{6}}{4} = \sin(285)\]
Easy to compute \[-\frac{\sqrt{6} + \sqrt{2}}{4}\] from a graph? You must be pretty badass with graphs.
lol, yes by drawing the graph. You know turning the lines in circles we have 90 -180 360 -270 he'll easily figure out angle by then :)
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 :)?
The question specifically asks to use the addition formulae, and besides, no 'circle' will give you the angle 285.
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 ? :)
sstarica, please, draw me a circle which can give the exact value I calculated without, in essence, just using the addition formulae in diagram form.
I can't draw it here, but you can twirl the line till you get the given angle, then you can figure out which angles to add. Those angles will be easy to compute since they're small, and in the end you can just add them up to find the sin of them ^_^
stop fighting guys
Oh, adding angles? Weird, just what I did. BTW drawing is harder than saying 285 = 360 - 75 => we need -sin(75)
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
we all make mistakes Inew :) even I, even the most intelligent, but noone said you're wrong, but your way is complicated.
try to explain it in a simple way :)
I was doing one half of it. See if you can apply the same logic to the next part - you might find it simlpifies to.. something nice.
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
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Alright we have :\[\cos(x)/\sec(x) + \sin(x)/\csc(x)\] sec x = 1/cosx and csc x = 1/sin x so: \[[\cos(x)/1/\cos(x)] + [\sin(x)/1/\sin(x)]\] simplify and you'll get : \[= \cos^2(x) +\sin^2(x)\] \[= 1\] Correct me if I'm wrong please ^_^
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.
It's alright, we have : \[=\cos^2(x)/sinx + 1/sinx\] \[= [\cos^2x + 1]/\sin(x)\] \[= \sin^2x/sinx = \sin(x)\] ^_^
ORLY?
maybe? :) correct me lol
it's not correct
\[\cos^2x + 1 \not= \sin^2x\]
hmm, where is my mistake
LOL! right >_< sorry
For what it's worth, I don't think it simplifies much nicer than how it starts, so is a bad question.
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
\[=\frac{\cos^2x}{\sin(x)} + 1/cos(x)\]\[= \frac{\cos(x)(cos^2x+1)}{\sin(x)cos(x)}\]\[= \frac{2-\sin^2x}{\sin(x)}\]\[= \frac{2}{\sin(x)} -\sin(x) = 2\csc(x) -\sin(x)\]
I'm not sure though, I was laggin ^_^|"
just go over it please :)
nope
wrong? ._.
yep
blah~, I'll try again ^_^
For the record, loco, my answer (sec(x)+cosec(x) - sin(x)) is right, but probably thr wrong form - it's hideous.
sec(x) + cos(x)cot(x), maybe?
don't know, my school's hw program sucks nuts
i only get 3 tries per problem
That program is retarded, then. Both of the answers I've given are right, but if the form matters then both could be shown as wrong. FFFFUUUUUUUUUUUUU
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
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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? :)
The answer =sec x + csc x - sin x =sec x + cos x cot x Now guess which is the required form :D Yes, got it above.
yeah he did or she
Oh, I'm a she. Hence the Newton name and everythi-- oh, wait a second ...
lol, then loco go with it, I got the same answer too ^_^, damn my body temperature increased, too much concentration ^^"
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
Oh, I'm english, so these american maths terms mean nothing to me ¬_¬ And trust me, that wasn't me trying.
american ed system succks retricemate
Don't worry, the person on the other thread got it wrong.
bloody teachers
yeah he did
lol, relax, so what if the teaher sucks? depend on yourself and trust your abilities :), I had the worse teachers, so I had to depend on myself and look where I've reached :) good luck loco ^_^
THANKS SSTARICA :)
for my own personal interest, where have you reaced?
np :)
what do u mean
"Look where I've reached" - I thought that meant a title/place, but now I think about I think it just means an ability level. My bad.
i think it can mean both. depends on what you're talking about

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