Question

what are the steps to calculate sin20 /cos70

Answer 1

without a calculator, you mean?

Answer 2

yes

Answer 3

gve me the steps?

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

No.. :(

Answer 11

lol...thx

Answer 12

so wats the answer

Answer 13

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

Answer 14

Do you think you could answer it?

Answer 15

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

Answer 16

I used the formula that Newton wrote above to show that sin(20) is the same as cos(70), therefore,
\[{\sin(20) \over \cos(70)}={\cos(70) \over \cos(70)}=1\]

Answer 17

What was the problem?

Answer 18

ok...thx again

Answer 19

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

Answer 20

You're welcome leonasmart!

Answer 21

Yeah I guess.

Answer 22

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

Answer 23

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

Answer 24

OK.. We are around.

Answer 25

ok.....go on INewton

Answer 26

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

Answer 27

:D I do care

Answer 28

OK WHAT about sin 30/cos 60

Answer 29

You do?

Answer 30

the same as the old one.

Answer 31

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.

Answer 32

OK.. I won't cheat :)

Answer 33

ok thx

Answer 34

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

Answer 35

Ok then.

Answer 36

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

Answer 37

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

Answer 38

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.

Answer 39

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.

Answer 40

No more hints Please!

Answer 41

There are radicals in the answer :/ Sorry.

Answer 42

Oh really?

Answer 43

kk

Answer 44

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.

Answer 45

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

Answer 46

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

Answer 47

Hmm, your equations are slightly inaccurate.

Answer 48

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

Answer 49

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

Answer 50

hmm ok.

While we're chatting, can I ask you a question: Just HOW good at Maths is Lokisan? If you can answer that.

Answer 51

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.

Answer 52

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.

Answer 53

What about we two make a change?

Answer 54

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 :(

Answer 55

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

Answer 56

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).

Answer 57

May I ask how old you are?

Answer 58

19. Yourself?

Answer 59

21

Answer 60

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

Answer 61

It's right

Answer 62

x+y=30/35?

Answer 63

Yep

Answer 64

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

Answer 65

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

Answer 66

yeah, sure (it's boring from there)

Answer 67

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

Answer 68

Obviously I did. :(

Answer 69

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

Answer 70

You know I will get two values for x and y, substitute them in two of the four equations to get a and b.

Answer 71

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

Answer 72

I actually have a nice question I just found, but I won't have been able to type it up for ~ 10/15 minutes

Answer 73

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)

Answer 74

Waiting for it :)

Answer 75

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

Answer 76

Exactly. I guess it's longer than most though. And most people here do problems to help people more than for just doing them.

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

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.

Answer 83

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 \[\int^2_{\frac{1}{2}} \frac{\sin x}{x(\sin x + \sin 1/x)} \]

Answer 84

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

Answer 85

There is a typo. at least one.

Answer 86

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

Answer 87

2nd one

Answer 88

Ah, an extra = sign :/ oops

Answer 89

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

Answer 90

For which part?

Answer 91

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

Answer 92

is the substitution a-x?

Answer 93

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.

Answer 94

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

Answer 95

whose good i writing essays

Answer 96

I am, but not for you, Mary.

Answer 97

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

Answer 98

can you help me write it