anonymous
  • anonymous
Use an addition or subtraction formula to find the exact value of the expression. sin(285)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Well, lets assume you know the 'obvious' values of sin - {0, 30, 45, 60, 90} + 360n. How can you use them to make 285?
anonymous
  • anonymous
add them?
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so what do i add?
anonymous
  • anonymous
That's where i'm cofused
anonymous
  • anonymous
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
anonymous
  • anonymous
¬_¬ Sin 150 = Sin 30 Sin 135 = Sin 45
anonymous
  • anonymous
I do not get it.
anonymous
  • anonymous
Do you not get why that is true?
anonymous
  • anonymous
Or do you not get that sin(150 + 135) = sin 285
anonymous
  • anonymous
both
anonymous
  • anonymous
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.
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
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.
anonymous
  • anonymous
-90?
anonymous
  • anonymous
Last line is -sin(75). Where did you get 90 from?
anonymous
  • anonymous
i have no idea what -sin 35= ?
anonymous
  • anonymous
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)
anonymous
  • anonymous
the answer is?
anonymous
  • anonymous
hmm, INew, wouldn't it be easier if you just drew the graph and compute it from the graph? :) it'll be alot easier ^_^
anonymous
  • anonymous
sin(45) = cos(45) = 1/sqrt(2) sin(30) = 1/2 cos(30) = sqrt(3)/2 \[\implies -\sin(75) = -\frac{1}{\sqrt{2}} \cdot \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = -\frac{1+ \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{2} + \sqrt{6}}{4} = \sin(285)\]
anonymous
  • anonymous
Easy to compute \[-\frac{\sqrt{6} + \sqrt{2}}{4}\] from a graph? You must be pretty badass with graphs.
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
any *
anonymous
  • anonymous
Just admit you were wrong, please, and I won't have to own you.
anonymous
  • anonymous
lol, own who?
anonymous
  • anonymous
the graph is actually easier, never realized it :p
anonymous
  • anonymous
LOL, do you know how yo use it loco :)?
anonymous
  • anonymous
The question specifically asks to use the addition formulae, and besides, no 'circle' will give you the angle 285.
anonymous
  • anonymous
lol, yes it will :) , you've never used it right?
anonymous
  • anonymous
kind of, ta showed it to us. still iffy on it but practicing.
anonymous
  • anonymous
besides loco, the information INew has given you also is valuable, work with both ^_^
anonymous
  • anonymous
thanks guys :)
anonymous
  • anonymous
np ^_^
anonymous
  • anonymous
but did you figure it out ? :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 ^_^
anonymous
  • anonymous
stop fighting guys
anonymous
  • anonymous
Oh, adding angles? Weird, just what I did. BTW drawing is harder than saying 285 = 360 - 75 => we need -sin(75)
anonymous
  • anonymous
Simplify the trigonometric expression. cos[x]/sec[x] + sin[x]/csc[x]
anonymous
  • anonymous
Sorry, I thought they were implying they could do the hard part in a circle, not the easy part :/
anonymous
  • anonymous
I'm not fighting LOL, we're discussing loco ^_^
anonymous
  • anonymous
1/sec = cos. 1/csc = sin. cos * 1/sec = .... etc
anonymous
  • anonymous
how about we i mean u guys solve this problem
anonymous
  • anonymous
you still didn't solve it? .-.
anonymous
  • anonymous
no, i hate this stuff
anonymous
  • anonymous
don't say that, you still didn't master it , but you will with practice and it'll be easy on you ^_^
anonymous
  • anonymous
i've been practicing a lot, but still having trouble :'(
anonymous
  • anonymous
\[\frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2x\]
anonymous
  • anonymous
it's alright, keep on practicing and you'll master it :)
anonymous
  • anonymous
sstarica help?
anonymous
  • anonymous
hold on :)
anonymous
  • anonymous
newton it's wrong
anonymous
  • anonymous
What's wrong? (Before answering, note that I am never wrong)
anonymous
  • anonymous
sorry, i thought that was the answer
anonymous
  • anonymous
we all make mistakes Inew :) even I, even the most intelligent, but noone said you're wrong, but your way is complicated.
anonymous
  • anonymous
try to explain it in a simple way :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
He meant I am wrong about the second problem he posted in here. I only did half of it
anonymous
  • anonymous
Yes, my FIRST method was comlpicated, but my second was amazing.
anonymous
  • anonymous
he posted another question?
anonymous
  • anonymous
lol, good that you've made it easier ^_^
anonymous
  • anonymous
Thankfully still no graphs required
anonymous
  • anonymous
oh that one , LOL LOCO! you've posted it in a different thread =D and I was writing it down
anonymous
  • anonymous
u can write it here
anonymous
  • anonymous
nvm it, INew answered it though :)
anonymous
  • anonymous
it's not right though
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
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 ^_^
anonymous
  • anonymous
you are right :) i got the same answer
anonymous
  • anonymous
No, you aren't wrong. Unless you count not using \frac{}{} for fractions as wrong (I personally do)
anonymous
  • anonymous
well done ^_^
anonymous
  • anonymous
lol, I didn't get you INew
anonymous
  • anonymous
\[\frac{1}{2} \geq 1/2\]
anonymous
  • anonymous
hmm, I can't find the frac() symbol lol >_<
anonymous
  • anonymous
still don't know how to make that fraction line ^_^"
anonymous
  • anonymous
last problem.... Simplify the trigonometric expression. cos[x]/ tan[x] + sex[x]
anonymous
  • anonymous
frac{a}{b} gives\[\frac{a}{b}\]
anonymous
  • anonymous
oh, thank you :)
anonymous
  • anonymous
sex(x) - lolololololol
anonymous
  • anonymous
LOL
anonymous
  • anonymous
sec[x], sorry :P
anonymous
  • anonymous
wish math was that fun
anonymous
  • anonymous
I general, always write tan as sin/cos , sec as 1/cos etc.
anonymous
  • anonymous
When trying to simplify things.
anonymous
  • anonymous
It's alright, we have : \[=\cos^2(x)/sinx + 1/sinx\] \[= [\cos^2x + 1]/\sin(x)\] \[= \sin^2x/sinx = \sin(x)\] ^_^
anonymous
  • anonymous
ORLY?
anonymous
  • anonymous
maybe? :) correct me lol
anonymous
  • anonymous
it's not correct
anonymous
  • anonymous
\[\cos^2x + 1 \not= \sin^2x\]
anonymous
  • anonymous
hmm, where is my mistake
anonymous
  • anonymous
LOL! right >_< sorry
anonymous
  • anonymous
For what it's worth, I don't think it simplifies much nicer than how it starts, so is a bad question.
anonymous
  • anonymous
hmm, solve it :)
anonymous
  • anonymous
man, u guys are at it.
anonymous
  • anonymous
LOL, loco we're not fighting =P, hold on :)
anonymous
  • anonymous
Ugh, it doesn't simplify.. at all. But also sec x = 1/cos(x) ... >_> sorry
anonymous
  • anonymous
\[\sec x + \csc x -\sin x\] .. but I'm tired.
anonymous
  • anonymous
Loco left is for a new thread. Un coooooooooooool
anonymous
  • anonymous
i posted it in a new thread also, but u guys were to busy
anonymous
  • anonymous
2nd one
anonymous
  • anonymous
Was my answer right?
anonymous
  • anonymous
hold on
anonymous
  • anonymous
Joke, I know it was. Just unsure what the 'expected' form is.
anonymous
  • anonymous
it's wrong newton
anonymous
  • anonymous
\[=\frac{\cos^2x}{\sin(x)} + 1/cos(x)\]\[= \frac{\cos(x)(cos^2x+1)}{\sin(x)cos(x)}\]\[= \frac{2-\sin^2x}{\sin(x)}\]\[= \frac{2}{\sin(x)} -\sin(x) = 2\csc(x) -\sin(x)\]
anonymous
  • anonymous
I'm not sure though, I was laggin ^_^|"
anonymous
  • anonymous
just go over it please :)
anonymous
  • anonymous
nope
anonymous
  • anonymous
wrong? ._.
anonymous
  • anonymous
yep
anonymous
  • anonymous
blah~, I'll try again ^_^
anonymous
  • anonymous
For the record, loco, my answer (sec(x)+cosec(x) - sin(x)) is right, but probably thr wrong form - it's hideous.
anonymous
  • anonymous
sec(x) + cos(x)cot(x), maybe?
anonymous
  • anonymous
don't know, my school's hw program sucks nuts
anonymous
  • anonymous
i only get 3 tries per problem
anonymous
  • anonymous
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
anonymous
  • anonymous
at least it's not timed, i've been doing this since 9pm
anonymous
  • anonymous
it really pisses me off and the professor is a b**** and doesn't do anything about it
anonymous
  • anonymous
lol, both of you calm down and just try again ^_^
anonymous
  • anonymous
\[\frac{\cos x}{\tan x} + \sec x = \frac{\cos^2x}{\sin x} + \sec x = \cos x\cot x + \sec x = ?\]
anonymous
  • anonymous
Which can also be written as sec x + csc x - sin x , and many other things.
anonymous
  • anonymous
i'll wait for sstarica's answer
anonymous
  • anonymous
True, she (I think, but maybe he?) does have more fans than me, so that's a good call.
anonymous
  • anonymous
I'm a she =P
anonymous
  • anonymous
I'm a she =P
anonymous
  • anonymous
r u hot? ;)
anonymous
  • anonymous
just to make sure, what is the question loco? :)
anonymous
  • anonymous
let me find it
anonymous
  • anonymous
Simplify the trigonometric expression. cos[x]/ tan[x] + sec[x]
anonymous
  • anonymous
what on earth? lol what kind of quetion is that? >_>
anonymous
  • anonymous
idk.... curious loco mind
anonymous
  • anonymous
Don't worry about loco the sexpest . But in answer to the question of course she is, she does Maths.
anonymous
  • anonymous
ok... lets continue with math. i don't wanna go to jail
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I got : cscx -sinx + secx *sigh* not sure though, but that's what I got
anonymous
  • anonymous
cos^2x + 1 ---------- sinx
anonymous
  • anonymous
Blexting is about to say : This problem is trivial via the use of synthetic division.
anonymous
  • anonymous
Blexting sec x is not 1 / sin x, btw.
anonymous
  • anonymous
it's wrong blexting
anonymous
  • anonymous
lol
anonymous
  • anonymous
check mine please ^^" if I'm wrong, then I'm sorry
anonymous
  • anonymous
My answer (which sstarica now has, too) has been right all along.
anonymous
  • anonymous
ok this the last try i have left any final answers
anonymous
  • anonymous
lol INew, you got the same answer? :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
yeah he did or she
anonymous
  • anonymous
Oh, I'm a she. Hence the Newton name and everythi-- oh, wait a second ...
anonymous
  • anonymous
lol, then loco go with it, I got the same answer too ^_^, damn my body temperature increased, too much concentration ^^"
anonymous
  • anonymous
good question ._.
anonymous
  • anonymous
nope it didn't work. sorry guys. told u the system is bs
anonymous
  • anonymous
=-=......I'm sorry
anonymous
  • anonymous
i got a 90 on the quiz though
anonymous
  • anonymous
excellent :)
anonymous
  • anonymous
Out of interest, what level is the quiz? As in, what educational level?
anonymous
  • anonymous
THANKS FOR TRYING NEWTON (HE OR SHE?) AND SSTARICA (SEND ME UR # ;p)
anonymous
  • anonymous
PRE-CALC
anonymous
  • anonymous
lok, someone answered this question in the other thread >_< LOL!
anonymous
  • anonymous
but the professor is really tough, doesn't curve or anything
anonymous
  • anonymous
Oh, I'm english, so these american maths terms mean nothing to me ¬_¬ And trust me, that wasn't me trying.
anonymous
  • anonymous
american ed system succks retricemate
anonymous
  • anonymous
Don't worry, the person on the other thread got it wrong.
anonymous
  • anonymous
bloody teachers
anonymous
  • anonymous
yeah he did
anonymous
  • anonymous
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 ^_^
anonymous
  • anonymous
THANKS SSTARICA :)
anonymous
  • anonymous
for my own personal interest, where have you reaced?
anonymous
  • anonymous
np :)
anonymous
  • anonymous
what do u mean
anonymous
  • anonymous
"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.
anonymous
  • anonymous
i think it can mean both. depends on what you're talking about

Looking for something else?

Not the answer you are looking for? Search for more explanations.