what are the steps to calculate sin20 /cos70

- anonymous

what are the steps to calculate sin20 /cos70

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- schrodinger

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- anonymous

without a calculator, you mean?

- anonymous

yes

- anonymous

gve me the steps?

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## More answers

- 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

Glad, too, as 20 and 70 aren't the easier to calculate directly (but it can be done without THAT much hastle)

- anonymous

sin(20)=cos(90-20)=cos(70).
it's pretty obvious now, isn't it?

- anonymous

It was obvious before that, tbh. I could just sense it.

- anonymous

LOL!! I am sure it was obvious for you.. I was talking to leonasmart!

- anonymous

You're looking for a fight, aren't you Michael?

- anonymous

No.. :(

- anonymous

lol...thx

- anonymous

so wats the answer

- anonymous

Michael, What about you post a real challenging problem if you have one?

- anonymous

Do you think you could answer it?

- anonymous

I posted one of my 'fun' problems yesterday, btw. No one got it, so I had to give the solution.

- 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

What was the problem?

- anonymous

ok...thx again

- anonymous

Oh, nothing too serious, just some simultaneous equations - is it possible to link to old topics?

- anonymous

You're welcome leonasmart!

- anonymous

Yeah I guess.

- anonymous

i am going to try do one on my own now and ask u if it is correct

- anonymous

btw leonas if you want proof of that sin(90-x) thing consider a RA triangle.

- anonymous

OK.. We are around.

- anonymous

ok.....go on INewton

- anonymous

Congrats for the new title Newton<< (like you would care :P)

- anonymous

:D I do care

- anonymous

OK WHAT about sin 30/cos 60

- anonymous

You do?

- anonymous

the same as the old one.

- 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

OK.. I won't cheat :)

- anonymous

ok thx

- anonymous

if I got the answer, I would post it here..K?

- anonymous

Ok then.

- 2bornot2b

Can you please help me with my problem at
http://openstudy.com/groups/mathematics#/updates/4dadd133d6938b0b6b21a94d

- anonymous

about the hint you gave, what if x+y is 0?

- 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

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

No more hints Please!

- anonymous

There are radicals in the answer :/ Sorry.

- anonymous

Oh really?

- anonymous

kk

- 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

So far, I have the two equations, which can find x and y:
x+y=48/35 (1) and xy=9/35.

- anonymous

Yes, I'm not a big fan of irrational numbers either.

- anonymous

Hmm, your equations are slightly inaccurate.

- anonymous

But you are meant to obtain values for (x+y) and xy first, so your method is obviously right.

- anonymous

I know my method is right :P .. I want my answer to be right!! just gimme a minute to check my calculations.

- 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

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

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

What about we two make a change?

- 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

You didn't tell me what you think about lucas?

- 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

May I ask how old you are?

- anonymous

19. Yourself?

- anonymous

21

- anonymous

xy=3/35.. this can't be wrong.

- anonymous

It's right

- anonymous

x+y=30/35?

- anonymous

Yep

- anonymous

35xy=3 and 35x+35y=30 implies x(30-35x)=3

- anonymous

Am I allowed to use software to solve the quadratic equation? :(

- anonymous

yeah, sure (it's boring from there)

- anonymous

Really? Did I already get all the fun? :(

- anonymous

Obviously I did. :(

- anonymous

:( I'll keep a lookout for a nicer one

- 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

Yeah (x/y and a/b are interchangeable now, except their signs compliment each other)

- 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

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

Waiting for it :)

- anonymous

I wonder why nobody solved this problem. It's not that difficult.

- 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

Yeah. I usually don't solve problems that take too much time.

- anonymous

What major are you taking? ( assuming you already go to college )

- anonymous

Well University, but in England, so no 'major' - it's a slightly different system - just Maths.

- anonymous

:O you don't have majors in England.. So you're going for a degree in Mathematics.

- anonymous

Indeed. Ugh I have the question but there is a problem in the LaTeX. Trying to find it.

- 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

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

OK. I think that's right but if I see a typo I'll let you know

- anonymous

There is a typo. at least one.

- anonymous

Where? The last integral need a dx, but apart from that..?

- anonymous

2nd one

- anonymous

Ah, an extra = sign :/ oops

- anonymous

Are you sure about the second integral? I think something is wrong.

- anonymous

For which part?

- anonymous

f(a-x) / [(f(x) + f(a-x)] ?

- anonymous

is the substitution a-x?

- 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

Oh Yeah.. That's what I was stuck at :(

- anonymous

whose good i writing essays

- anonymous

I am, but not for you, Mary.

- anonymous

Hint, though Mary: It's ' Who's' (short for who is) not whose.

- anonymous

can you help me write it

- anonymous

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

- 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

mary try to find amistre64.. I am sure he will help you :)

- anonymous

xD so true

- anonymous

Okay.. We will see

- anonymous

My mom is calling :)

- anonymous

Does that mean you are going to stop :(

- anonymous

I'll BRB

- anonymous

Ah, OK.

- anonymous

i newton plz help me go to the english writing center plz

- anonymous

Back.. I will start with the first one.

- 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

And THEN do the next integrals

- 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

HAHA.. I should use the definition of derivatives?

- anonymous

Erm, no? I can give you a (sort of big) hint if you want.

- anonymous

No.. Thanks.. I am just a bit sleepy. Can I leave it to tomorrow?

- 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

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

wait!

- anonymous

:( OK

- anonymous

This image is the QUESTION (formatted nicely) NOT the solution.

##### 1 Attachment

- 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

OH I have a quiz tomorrow that I didn't prepare for at all :(.. Wish me luck.
PEACE!!

- anonymous

Yeah, OK, bye.
Haha, good luck.

- anonymous

If you want it: See attached image

##### 1 Attachment

- anonymous

Hey I did it without looking at your answer. I think it's the right answer :)

- 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

To be honest, I didn't expect such a nice answer.

- anonymous

I will do the integrals now!!

- 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

I almost wrote the whole solution of the 2nd one but accidentally lost it :( .. I will write it back in few minutes.

- 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

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

Very nice.

- 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

OH you came!! Hello :)

- anonymous

LOL.. I feel SO dumb :( .. I complicated it too much

- anonymous

Hey (sorry was away again)

- 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

Good :D
But don't worry, these questions aren't meant to be easy

- 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

:DDD

- 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

Slight error:.
ln(a) and ln(1/a) are not in the same log so combine differently (easieR)

- anonymous

Just direct substitution with a=2 to get the value of the last integral :)

- anonymous

OH :(

- anonymous

You're right!! :(

- anonymous

But other than that yeah

- anonymous

Stupid mistake I know.

- anonymous

I hope I won't kill myself today!!

- anonymous

I=ln IaI?

- anonymous

Yep

- anonymous

Thank you so much.. This Question was so amazing. I liked it and I hated myself :P

- anonymous

Haha, you're welcome. I'll probably find some more sometime, but have to do some other stuff first today

- 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

has just started* right?

- 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:

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- anonymous

Bye!

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