anonymous
  • anonymous
what are the steps to calculate sin20 /cos70
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
without a calculator, you mean?
anonymous
  • anonymous
yes
anonymous
  • anonymous
gve me the steps?

Looking for something else?

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

More answers

anonymous
  • anonymous
Oh, 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
  • anonymous
Glad, too, as 20 and 70 aren't the easier to calculate directly (but it can be done without THAT much hastle)
anonymous
  • anonymous
sin(20)=cos(90-20)=cos(70). it's pretty obvious now, isn't it?
anonymous
  • anonymous
It was obvious before that, tbh. I could just sense it.
anonymous
  • anonymous
LOL!! I am sure it was obvious for you.. I was talking to leonasmart!
anonymous
  • anonymous
You're looking for a fight, aren't you Michael?
anonymous
  • anonymous
No.. :(
anonymous
  • anonymous
lol...thx
anonymous
  • anonymous
so wats the answer
anonymous
  • anonymous
Michael, What about you post a real challenging problem if you have one?
anonymous
  • anonymous
Do you think you could answer it?
anonymous
  • anonymous
I posted one of my 'fun' problems yesterday, btw. No one got it, so I had to give the solution.
anonymous
  • anonymous
I 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
  • anonymous
What was the problem?
anonymous
  • anonymous
ok...thx again
anonymous
  • anonymous
Oh, nothing too serious, just some simultaneous equations - is it possible to link to old topics?
anonymous
  • anonymous
You're welcome leonasmart!
anonymous
  • anonymous
Yeah I guess.
anonymous
  • anonymous
i am going to try do one on my own now and ask u if it is correct
anonymous
  • anonymous
btw leonas if you want proof of that sin(90-x) thing consider a RA triangle.
anonymous
  • anonymous
OK.. We are around.
anonymous
  • anonymous
ok.....go on INewton
anonymous
  • anonymous
Congrats for the new title Newton<< (like you would care :P)
anonymous
  • anonymous
:D I do care
anonymous
  • anonymous
OK WHAT about sin 30/cos 60
anonymous
  • anonymous
You do?
anonymous
  • anonymous
the same as the old one.
anonymous
  • anonymous
my 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
  • anonymous
OK.. I won't cheat :)
anonymous
  • anonymous
ok thx
anonymous
  • anonymous
if I got the answer, I would post it here..K?
anonymous
  • anonymous
Ok then.
2bornot2b
  • 2bornot2b
anonymous
  • anonymous
about the hint you gave, what if x+y is 0?
anonymous
  • anonymous
It 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
  • anonymous
I 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
  • anonymous
No more hints Please!
anonymous
  • anonymous
There are radicals in the answer :/ Sorry.
anonymous
  • anonymous
Oh really?
anonymous
  • anonymous
kk
anonymous
  • anonymous
Yes. 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
  • anonymous
So far, I have the two equations, which can find x and y: x+y=48/35 (1) and xy=9/35.
anonymous
  • anonymous
Yes, I'm not a big fan of irrational numbers either.
anonymous
  • anonymous
Hmm, your equations are slightly inaccurate.
anonymous
  • anonymous
But you are meant to obtain values for (x+y) and xy first, so your method is obviously right.
anonymous
  • anonymous
I know my method is right :P .. I want my answer to be right!! just gimme a minute to check my calculations.
anonymous
  • anonymous
hmm ok. While we're chatting, can I ask you a question: Just HOW good at Maths is Lokisan? If you can answer that.
anonymous
  • anonymous
Well 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
  • anonymous
I 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
  • anonymous
What about we two make a change?
anonymous
  • anonymous
Haha, 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
  • anonymous
You didn't tell me what you think about lucas?
anonymous
  • anonymous
Oh, 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
  • anonymous
May I ask how old you are?
anonymous
  • anonymous
19. Yourself?
anonymous
  • anonymous
21
anonymous
  • anonymous
xy=3/35.. this can't be wrong.
anonymous
  • anonymous
It's right
anonymous
  • anonymous
x+y=30/35?
anonymous
  • anonymous
Yep
anonymous
  • anonymous
35xy=3 and 35x+35y=30 implies x(30-35x)=3
anonymous
  • anonymous
Am I allowed to use software to solve the quadratic equation? :(
anonymous
  • anonymous
yeah, sure (it's boring from there)
anonymous
  • anonymous
Really? Did I already get all the fun? :(
anonymous
  • anonymous
Obviously I did. :(
anonymous
  • anonymous
:( I'll keep a lookout for a nicer one
anonymous
  • anonymous
You know I will get two values for x and y, substitute them in two of the four equations to get a and b.
anonymous
  • anonymous
Yeah (x/y and a/b are interchangeable now, except their signs compliment each other)
anonymous
  • anonymous
I actually have a nice question I just found, but I won't have been able to type it up for ~ 10/15 minutes
anonymous
  • anonymous
This 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
  • anonymous
Waiting for it :)
anonymous
  • anonymous
I wonder why nobody solved this problem. It's not that difficult.
anonymous
  • anonymous
Exactly. I guess it's longer than most though. And most people here do problems to help people more than for just doing them.
anonymous
  • anonymous
Yeah. I usually don't solve problems that take too much time.
anonymous
  • anonymous
What major are you taking? ( assuming you already go to college )
anonymous
  • anonymous
Well University, but in England, so no 'major' - it's a slightly different system - just Maths.
anonymous
  • anonymous
:O you don't have majors in England.. So you're going for a degree in Mathematics.
anonymous
  • anonymous
Indeed. Ugh I have the question but there is a problem in the LaTeX. Trying to find it.
anonymous
  • anonymous
It'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
  • anonymous
Let \[I = \int^a_0 \frac{f(x)}{f(x)+f(a-x)} \mathbb{d}x\] Use a substitution to show that \[I = \int^a_0 = \frac{f(a-x)}{f(x)+f(a-x)}\mathbb{d}x \] and hence evaluate I in terms of a. Use this result to evaluate the integrals \[\int^1_0 \frac{ln(x+1)}{ln(2+x-x^2)}\mathbb{d}x\ \] 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
  • anonymous
OK. I think that's right but if I see a typo I'll let you know
anonymous
  • anonymous
There is a typo. at least one.
anonymous
  • anonymous
Where? The last integral need a dx, but apart from that..?
anonymous
  • anonymous
2nd one
anonymous
  • anonymous
Ah, an extra = sign :/ oops
anonymous
  • anonymous
Are you sure about the second integral? I think something is wrong.
anonymous
  • anonymous
For which part?
anonymous
  • anonymous
f(a-x) / [(f(x) + f(a-x)] ?
anonymous
  • anonymous
is the substitution a-x?
anonymous
  • anonymous
Yes, u = a-x (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
  • anonymous
Oh Yeah.. That's what I was stuck at :(
anonymous
  • anonymous
whose good i writing essays
anonymous
  • anonymous
I am, but not for you, Mary.
anonymous
  • anonymous
Hint, though Mary: It's ' Who's' (short for who is) not whose.
anonymous
  • anonymous
can you help me write it
anonymous
  • anonymous
plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
anonymous
  • anonymous
Here is the proof quickly: \[y=a-x \implies dy=-dx; I=-\int\limits_{a}^{0}{f(a-y) \over f(y)+f(a-y)}dy\] \[\implies I=\int\limits_{0}^{a}{f(a-y) \over f(y)+f(a-y)}dy\]
anonymous
  • anonymous
Perfect. I hope you like how you use it for the next bit :D I did.
anonymous
  • anonymous
mary try to find amistre64.. I am sure he will help you :)
anonymous
  • anonymous
xD so true
anonymous
  • anonymous
Okay.. We will see
anonymous
  • anonymous
My mom is calling :)
anonymous
  • anonymous
Does that mean you are going to stop :(
anonymous
  • anonymous
I'll BRB
anonymous
  • anonymous
Ah, OK.
anonymous
  • anonymous
i newton plz help me go to the english writing center plz
anonymous
  • anonymous
Back.. I will start with the first one.
anonymous
  • anonymous
OK. you should probably consider the part 'and hence evaluate I in terms of a' (after the bit you did - it's very useful.
anonymous
  • anonymous
And THEN do the next integrals
anonymous
  • anonymous
Hello? 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
  • anonymous
HAHA.. I should use the definition of derivatives?
anonymous
  • anonymous
Erm, no? I can give you a (sort of big) hint if you want.
anonymous
  • anonymous
No.. Thanks.. I am just a bit sleepy. Can I leave it to tomorrow?
anonymous
  • anonymous
OK 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
  • anonymous
I'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
  • anonymous
wait!
anonymous
  • anonymous
:( OK
anonymous
  • anonymous
This image is the QUESTION (formatted nicely) NOT the solution.
1 Attachment
anonymous
  • anonymous
Well, 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
  • anonymous
OH I have a quiz tomorrow that I didn't prepare for at all :(.. Wish me luck. PEACE!!
anonymous
  • anonymous
Yeah, OK, bye. Haha, good luck.
anonymous
  • anonymous
If you want it: See attached image
1 Attachment
anonymous
  • anonymous
Hey I did it without looking at your answer. I think it's the right answer :)
anonymous
  • anonymous
since \[\int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=\int\limits_{0}^{a}{f(a-x) \over f(x)+f(a-x)}dx\], then \[\int\limits_{0}^{a}{f(a-x)-f(a) \over f(x)+f(a-x)}dx=\int\limits_{0}^{a}(1-2{f(x) \over f(x)+f(a-x)})dx=0\] \[\implies a-2\int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=0 \] \[\implies \int\limits_{0}^{a}{f(x) \over f(x)+f(a-x)}dx=a/2\]
anonymous
  • anonymous
To be honest, I didn't expect such a nice answer.
anonymous
  • anonymous
I will do the integrals now!!
anonymous
  • anonymous
The first integral is very easy.. Call the integral I. \[I=\int\limits_{0}^{1}{\ln(x+1) \over \ln(2+x-x^2)}dx=\int\limits_{0}^{1}{\ln(x+1) \over \ln(x+1)+\ln(2-x)}dx\] let's check if it's in the form we're looking for: let f(x)=ln(x+1), a=1. Therefore f(a-x)=ln(1-x+1)=ln(2-x).. Hence I=a/2=0.5 :O
anonymous
  • anonymous
I almost wrote the whole solution of the 2nd one but accidentally lost it :( .. I will write it back in few minutes.
anonymous
  • anonymous
Here 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
  • anonymous
This integral is in the form: \[\int\limits_{0}^{a}{f(a-x) \over f(x)+f(a-x)}dx\] 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
  • anonymous
Very nice.
anonymous
  • anonymous
The 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
  • anonymous
OH you came!! Hello :)
anonymous
  • anonymous
LOL.. I feel SO dumb :( .. I complicated it too much
anonymous
  • anonymous
Hey (sorry was away again)
anonymous
  • anonymous
It'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
  • anonymous
Good :D But don't worry, these questions aren't meant to be easy
anonymous
  • anonymous
I 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
  • anonymous
:DDD
anonymous
  • anonymous
Call that integral I. So: \[2I=\int\limits_{1/a}^{a}{1 \over x}dx=\left[ \ln \left| x \right| \right]_{1/a}^{a}=\ln \left| a-1/a \right| \implies I=\ln \sqrt{a-1/a}\]
anonymous
  • anonymous
Slight error:. ln(a) and ln(1/a) are not in the same log so combine differently (easieR)
anonymous
  • anonymous
Just direct substitution with a=2 to get the value of the last integral :)
anonymous
  • anonymous
OH :(
anonymous
  • anonymous
You're right!! :(
anonymous
  • anonymous
But other than that yeah
anonymous
  • anonymous
Stupid mistake I know.
anonymous
  • anonymous
I hope I won't kill myself today!!
anonymous
  • anonymous
I=ln IaI?
anonymous
  • anonymous
Yep
anonymous
  • anonymous
Thank you so much.. This Question was so amazing. I liked it and I hated myself :P
anonymous
  • anonymous
Haha, you're welcome. I'll probably find some more sometime, but have to do some other stuff first today
anonymous
  • anonymous
Good 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
  • anonymous
has just started* right?
anonymous
  • anonymous
Yes, 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:
1 Attachment
anonymous
  • anonymous
Bye!

Looking for something else?

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