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anonymous
 5 years ago
what are the steps to calculate sin20 /cos70
anonymous
 5 years ago
what are the steps to calculate sin20 /cos70

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0without a calculator, you mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, lol, I thought it would be some boring addition/subtraction formulae, but I forgot a key fact: sin(x) = cos(90  x) for all x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Glad, too, as 20 and 70 aren't the easier to calculate directly (but it can be done without THAT much hastle)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin(20)=cos(9020)=cos(70). it's pretty obvious now, isn't it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It was obvious before that, tbh. I could just sense it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL!! I am sure it was obvious for you.. I was talking to leonasmart!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're looking for a fight, aren't you Michael?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Michael, What about you post a real challenging problem if you have one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you think you could answer it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I posted one of my 'fun' problems yesterday, btw. No one got it, so I had to give the solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used the formula that Newton wrote above to show that sin(20) is the same as cos(70), therefore, \[{\sin(20) \over \cos(70)}={\cos(70) \over \cos(70)}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What was the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, nothing too serious, just some simultaneous equations  is it possible to link to old topics?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome leonasmart!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am going to try do one on my own now and ask u if it is correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw leonas if you want proof of that sin(90x) thing consider a RA triangle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Congrats for the new title Newton<< (like you would care :P)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK WHAT about sin 30/cos 60

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the same as the old one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my question (but answer is there so you can cheat :( ) http://openstudy.com/updates/4dac7055d6938b0b7bb9a74d#/updates/4dac7055d6938b0b7bb9a74d hope links work here. Note, I can ask some MUCH harder ones if you want.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK.. I won't cheat :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if I got the answer, I would post it here..K?

2bornot2b
 5 years ago
Best ResponseYou've already chosen the best response.0Can you please help me with my problem at http://openstudy.com/groups/mathematics#/updates/4dadd133d6938b0b6b21a94d

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0about the hint you gave, what if x+y is 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It makes no difference (but if they were the rest wouldn't fit anyway, so you can forget that possibility, as they cannot satisfy it). Another hint is to (if you realise why multiplying is helpful), to apply the same thing to the third one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I already did, and I almost got the values of x and y, but the numbers don't look right to me. They are a little bit complicated. So, I am doing it again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No more hints Please!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are radicals in the answer :/ Sorry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. The answers aren't nice, but the idea is good (most questions of the same type (same source) are much less dirty maths though, but it is probably one of the easiest to get into so I wanted to test the water with it). It's from a non calculator exam, though, so nothing TOO nasty.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So far, I have the two equations, which can find x and y: x+y=48/35 (1) and xy=9/35.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I'm not a big fan of irrational numbers either.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, your equations are slightly inaccurate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you are meant to obtain values for (x+y) and xy first, so your method is obviously right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know my method is right :P .. I want my answer to be right!! just gimme a minute to check my calculations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm ok. While we're chatting, can I ask you a question: Just HOW good at Maths is Lokisan? If you can answer that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well I don't know that much about him.. But, from what I have seen so far, I can say he's pretty good. You can't tell if he's special since there are not so many challenging problems here.. I think you would agree with that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I do agree with that, actually (mainly the reason I asked, as I have no way to find out). The only problem is no one on the site really wants challenging problems.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What about we two make a change?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha, I may post another 'fun' question later (but this time with nicer maths and not horrid answers). But unfortunately I don't think many others care :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You didn't tell me what you think about lucas?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I agree, he's obviously good. I pretty much think the same about not being able to judge on here though (he 'knows' higher level maths than me (I am much younger I am almost sure) but unsure exactly how good he is).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0May I ask how old you are?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0xy=3/35.. this can't be wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.035xy=3 and 35x+35y=30 implies x(3035x)=3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Am I allowed to use software to solve the quadratic equation? :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, sure (it's boring from there)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really? Did I already get all the fun? :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:( I'll keep a lookout for a nicer one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know I will get two values for x and y, substitute them in two of the four equations to get a and b.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah (x/y and a/b are interchangeable now, except their signs compliment each other)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I actually have a nice question I just found, but I won't have been able to type it up for ~ 10/15 minutes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is much nicer (it's more if you see the trick it's a very quick, nice problem, rather than grinding even after spotting how to do it)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wonder why nobody solved this problem. It's not that difficult.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. I guess it's longer than most though. And most people here do problems to help people more than for just doing them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. I usually don't solve problems that take too much time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What major are you taking? ( assuming you already go to college )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well University, but in England, so no 'major'  it's a slightly different system  just Maths.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:O you don't have majors in England.. So you're going for a degree in Mathematics.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed. Ugh I have the question but there is a problem in the LaTeX. Trying to find it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's 10:26 PM at where I am, and I didn't get enough sleep last night. So, if it would take me too long, I might delay solving it to tomorrow morning.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let \[I = \int^a_0 \frac{f(x)}{f(x)+f(ax)} \mathbb{d}x\] Use a substitution to show that \[I = \int^a_0 = \frac{f(ax)}{f(x)+f(ax)}\mathbb{d}x \] and hence evaluate I in terms of a. Use this result to evaluate the integrals \[\int^1_0 \frac{ln(x+1)}{ln(2+xx^2)}\mathbb{d}x\ \] and \[\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin (x + \pi/4)} \mathbb{d}x \] Evaluate \[\int^2_{\frac{1}{2}} \frac{\sin x}{x(\sin x + \sin 1/x)} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK. I think that's right but if I see a typo I'll let you know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is a typo. at least one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where? The last integral need a dx, but apart from that..?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, an extra = sign :/ oops

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you sure about the second integral? I think something is wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(ax) / [(f(x) + f(ax)] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the substitution ax?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, u = ax (or whatever you call the variable) Note u and x are just dummy variables, so if you get the same thing in u as the one in x it is the same integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh Yeah.. That's what I was stuck at :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whose good i writing essays

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am, but not for you, Mary.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hint, though Mary: It's ' Who's' (short for who is) not whose.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you help me write it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here is the proof quickly: \[y=ax \implies dy=dx; I=\int\limits_{a}^{0}{f(ay) \over f(y)+f(ay)}dy\] \[\implies I=\int\limits_{0}^{a}{f(ay) \over f(y)+f(ay)}dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perfect. I hope you like how you use it for the next bit :D I did.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mary try to find amistre64.. I am sure he will help you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that mean you are going to stop :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i newton plz help me go to the english writing center plz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Back.. I will start with the first one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK. you should probably consider the part 'and hence evaluate I in terms of a' (after the bit you did  it's very useful.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And THEN do the next integrals

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hello? I really hope you saw that (and get a general solution in terms of a for the integral), otherwise I think the integrals are pretty much impossible.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0HAHA.. I should use the definition of derivatives?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Erm, no? I can give you a (sort of big) hint if you want.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No.. Thanks.. I am just a bit sleepy. Can I leave it to tomorrow?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK sure. But (small hint but if you don't want one at all ignore if you can: you have two things from that are both the same integral. You have to use both of them, and you should be able to find something nicer. But cya

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll attach a solution (image) later in this thread if you wanna check when I'm not here, but don't feel like you have to look at it or anything.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This image is the QUESTION (formatted nicely) NOT the solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I will try it later. I will try my best not to see the solution :).. Thanks a lot anyway. It was nice talking to you. See you soon.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH I have a quiz tomorrow that I didn't prepare for at all :(.. Wish me luck. PEACE!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, OK, bye. Haha, good luck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you want it: See attached image

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey I did it without looking at your answer. I think it's the right answer :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since \[\int\limits_{0}^{a}{f(x) \over f(x)+f(ax)}dx=\int\limits_{0}^{a}{f(ax) \over f(x)+f(ax)}dx\], then \[\int\limits_{0}^{a}{f(ax)f(a) \over f(x)+f(ax)}dx=\int\limits_{0}^{a}(12{f(x) \over f(x)+f(ax)})dx=0\] \[\implies a2\int\limits_{0}^{a}{f(x) \over f(x)+f(ax)}dx=0 \] \[\implies \int\limits_{0}^{a}{f(x) \over f(x)+f(ax)}dx=a/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To be honest, I didn't expect such a nice answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will do the integrals now!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The first integral is very easy.. Call the integral I. \[I=\int\limits_{0}^{1}{\ln(x+1) \over \ln(2+xx^2)}dx=\int\limits_{0}^{1}{\ln(x+1) \over \ln(x+1)+\ln(2x)}dx\] let's check if it's in the form we're looking for: let f(x)=ln(x+1), a=1. Therefore f(ax)=ln(1x+1)=ln(2x).. Hence I=a/2=0.5 :O

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I almost wrote the whole solution of the 2nd one but accidentally lost it :( .. I will write it back in few minutes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here it is: \[\sin (x+\pi/4)=\sin x \cos(\pi/4)+\cos x (\sin \pi/4)={1 \over \sqrt2}(\sin x +\cos x)\] the integral can be written as: \[\int\limits_{0}^{\pi/2}{\sin x \over \sin x+\cos x}dx= \sqrt2\int\limits_{0}^{\pi/2} {\cos(\pi/2 x) \over \cos (\pi/2 x)+\cos x}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This integral is in the form: \[\int\limits_{0}^{a}{f(ax) \over f(x)+f(ax)}dx\] with f(x)=cos x and a=pi/2. Hence, \[\int\limits_{0}^{\pi/2}{\sin x \over \sin(x+\pi/4)}dx=\sqrt2 {(\pi/2) \over 2}=\sqrt 2 {\pi \over 4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The last part is a similar idea, but you don't use the previous result, but instead the previous general idea. But yeah, very nicely done. A slight note, the value of I (a/2) may be slightly easier proved by just added the two values for I  they total to 1. So: \[2I = \int^a_01\mathbb{d}x = [x]^a_0 = a \implies I = \frac{a}{2} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH you came!! Hello :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL.. I feel SO dumb :( .. I complicated it too much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey (sorry was away again)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's fine.. I still feel SO bad about taking too long in finding that I=a/2.. Anyway, I think I figured out how to do the last integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good :D But don't worry, these questions aren't meant to be easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I used a substitution (y=1/x) and proved that: \[\int\limits_{1/a}^{a}{f(x) \over x[f(x)+f(1/x)]}dx=\int\limits_{1/a}^{a}{f(1/x) \over x[f(x)+f(1/x)]}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Call that integral I. So: \[2I=\int\limits_{1/a}^{a}{1 \over x}dx=\left[ \ln \left x \right \right]_{1/a}^{a}=\ln \left a1/a \right \implies I=\ln \sqrt{a1/a}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Slight error:. ln(a) and ln(1/a) are not in the same log so combine differently (easieR)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just direct substitution with a=2 to get the value of the last integral :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But other than that yeah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Stupid mistake I know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hope I won't kill myself today!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you so much.. This Question was so amazing. I liked it and I hated myself :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha, you're welcome. I'll probably find some more sometime, but have to do some other stuff first today

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck in whatever you have to do (if you do believe in luck) :).. I will try to find some nice problems since my weekend just started and I have plenty of time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0has just started* right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, it is that. And actually, I just found another integration question you might like, but I have to go so won't be able to offer much help right now. See you on this thread later/tomorrow, though, probably. Or just around the site. Hope you enjoy:
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