provement

RauLiTo · 2006-03-12 23:10:24

prove it :
tan20 tan30 tan40 = tan10

sorry alot of questions these days

George,Y · 2006-03-13 02:42:42

i simplified it and finally made it through!

it really requires trig formula mastering

tan20 tan30 tan40=sin20 sin30 sin40 /( cos20 cos30 cos40) ; tan10=sin10/cos10=cos80/sin80

cos20 cos30 cos40 = (cos20 sin60) cos40 = 1/2 (sin80+sin40) cos40 =1/2 sin80 cos40 + 1/4 sin80

while
sin20 sin30 sin40 = (sin20 sin40) sin30 = 1/2 (cos20-cos60) sin30 = 1/2 cos20 sin30 - 1/2 sin30 sin30 = 1/4 (sin10+sin50) + 1/4 cos60-1/4 = 1/4 sin10 + 1/4cos40-1/8 = 1/4 sin10 + 1/4 (cos40-cos60) = 1/4 sin10 + 1/2 sin10 sin50=1/4 sin10 +1/2 sin10 cos40

so the left side=sin10/sin80=tan10=the right side

RauLiTo · 2006-03-13 04:33:55

thank you mr smaaaaaaaart ! thats really nice thank you again for helping me

George,Y · 2006-03-13 14:59:52

you r welcome

RauLiTo · 2006-03-17 23:07:59

can i get easir solvement for that ?
i think there is easier than this tall one

RauLiTo · 2006-03-18 03:59:30

no one can help me ? ganesh and the others help please

Jai Ganesh · 2006-03-18 04:18:37

I think
sin(a+b)=sinacosb+cosasinb and
cos(a+b)=cosacosb-sinasinb can be of some help. Please give me some time. I shall try and post the reply.

RauLiTo · 2006-03-18 23:39:36

Take your time ... i tried with it more than once but u know if it stucks ... !

Jai Ganesh · 2006-03-19 00:23:26

RauLiTo, I think these formulae should be helpful in solving the problem.

sin2A=2sinAcosA

cos2A=cos²A-sin²A

tan2A=2tanA/(1-tan²A)

sin3A=3sinA-4sin³A

cos3A=4cos³A-3cosA

tan3A=(3tanA-tan³A)/(1-3tan²A)

sin(A+B+C)=sinAcosBcosC+cosAsinBcosC+cosAcosBsinC-sinAsinBsinC

cos(A+B+C)=cosAcosBcosC-cosAsinBsinC-sinAcosBsinC-sinAsinBcosC

tan(A+B+C)=(tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB-tanBtanC-tanAtanC)

Jenilia · 2006-03-19 00:29:17

Well guys, I dont understand anything you are talking about, so I guess I will wait till it's time for me to learn. He He He!

Last edited by Jenilia (2006-03-19 00:29:58)

Jai Ganesh · 2006-03-19 02:29:20

RauLiTo,
tannθ=(S1-S3+s5-S7+....)/(1-S2+S4-S6+....)
where S1=ntanθ,
S2=nC2tan²θ,
S3=nC3tan³θ and so on.
The value of θ can be taken as 10 degrees,and using tan20=tan2(10), tan30=tan3(10) and tan40=tan4(10), I guess its possible to prove
tan20tan30tan40=tan10.

sameer mishra · 2010-09-02 21:17:46

if "3x+4y+5z=12"
then what is minimum valu of x^2+y^2+z^2

bobbym · 2010-09-03 01:25:15

Hi sameer mishra;

Have you covered lagrangian multipliers? You haven't shown me any work so I don't know what you know.

Bob · 2010-09-03 03:04:07

hi

3x + 4y + 5z = 12 is the equation of a plain in 3D and the other is the square of the distance from the origin to the plain. The minimum distance is the length of a line from the origin, that is perpendicular to the plain.

Do you know how to get the vector form for such a line ?

If not, I'll post it.

Bob

Bob · 2010-09-03 08:38:54

hi again,

to prove tan20.tan30.tan40 = tan10

thus putting (1) and (2) together

sameer mishra · 2010-09-03 21:53:58

what is area of triangle with sides 1-(x^1/2),5-(x^1/2),7-(x^1/2)
what is eaist method to solve it

bobbym · 2010-09-03 22:34:10

Hi sameer mishra;

if "3x+4y+5z=12"
then what is minimum valu of x^2+y^2+z^2

Are you not interested in the above question anymore?

what is area of triangle with sides 1-(x^1/2),5-(x^1/2),7-(x^1/2)
what is eaist method to solve it

You are not posting any of your work so I don't know where to try to help you.

You will need to solve for x first. Here you use the pythagorean theorem.

After some algebra:

Say √x = y and you get.

Can you solve this quadratic? Give it a try. Once you solve it I will show how to get the last step, the area!

Bob · 2010-09-04 07:51:29

Hi

There are too many threads muddled up here. Can someone split them please?

There is a formula for working out the area of any triangle (doesn't require a right angle)

where a, b, c are the lengths of the sides and

But you need to know x first, unless what you are wanting is a formula for the area in terms of x.

sameer mishra: You need to respond to the questions you have been asked about both of the problems you have posted.

Bob

bobbym · 2010-09-04 08:55:32

Hi bob and sameer;

There is a formula for working out the area of any triangle (doesn't require a right angle)

Good point. I figured he needed x so without assuming that triangle is right it is going to be difficult to get it.

There are too many threads muddled up here. Can someone split them please?

That would be something good but I don't have that capability here. It does get confusing but I can't at present do anything about it.

sameer mishra: You need to respond to the questions you have been asked about both of the problems you have posted.

Lately, and I am not saying this will happen for sure, there have been a rash of questions by posters whom never return. This is somewhat confusing to me. The negative side of me thinks that some posters now have no interest at all other than having someone, anyone, do their homework for them. I don't have to tell you, that is akin to cheating. They post the same problem on several forums and just wait. This is a wasteful use of the good people who donate their time. Some moderators elsewhere react strongly to that.

I remember about a week or so ago, a guy came in with 20 questions. He put guesses down for his work and answers and then requested we check them for him. That is an insult as well as being unethical. I only mention this because learn2teach has been discussing this with us.

Bob · 2010-09-04 20:23:52

hi bobbym,

I agree. I like helping people with their maths, but, if they're genuinely stuck on homework and wnat to learn more about the topic, then just a straight 'heres the answer' isn't going to do them any good. They won't be able to make a post when they're doing their exam!

I'd prefer it if someone said: 'Its my homework and I want to get the answer myself but please could someone point me in the right direction.'

It would also be courteous to say thanks afterwards. Or even some feedback that says whether the help provided actually clarified it for them. If someone posted back and said 'Im still a bit confused about ......'. No problem to add more help where it's genuinely needed.

For anyone else reading this, please take note!

Bob

ps. bobbym, I'm still working on the Cauchy problem. Thought I had it the other day, but I just went round in a perfect circle and landed back where I'd started.

pps D'ya think they think we get paid for this?

Last edited by Bob (2010-09-04 20:30:08)

bobbym · 2010-09-04 20:40:36

Hi bob;
Naturally I prefer people who are interested in math like I am. I am not too keen on assisting in the fooling of someone's teacher into thinking the person understands the material. I agree, doing that, does a lot of harm to the poster.

pps D'ya think they think we get paid for this?

Some forums pay their moderators. No math forums that I know of do that. Sports forums do that. All the people I know on forums do it because they want to.

George,Y · 2010-09-05 18:47:40

Can't believe I was so smart!
Now it is a little difficult to understand my own words!

sameer mishra · 2010-09-06 00:52:12

what is range of
F(x)=1/(3-5cosx)

sameer mishra · 2010-09-06 01:26:51

why sin30=1/2

sameer mishra · 2010-09-06 01:28:13

why sino=0

Math Is Fun Forum

#1 2006-03-12 23:10:24

provement

#2 2006-03-13 02:42:42

Re: provement

#3 2006-03-13 04:33:55

Re: provement

#4 2006-03-13 14:59:52

Re: provement

#5 2006-03-17 23:07:59

Re: provement

#6 2006-03-18 03:59:30

Re: provement

#7 2006-03-18 04:18:37

Re: provement

#8 2006-03-18 23:39:36

Re: provement

#9 2006-03-19 00:23:26

Re: provement

#10 2006-03-19 00:29:17

Re: provement

#11 2006-03-19 02:29:20

Re: provement

#12 2010-09-02 21:17:46

Re: provement

#13 2010-09-03 01:25:15

Re: provement

#14 2010-09-03 03:04:07

Re: provement

#15 2010-09-03 08:38:54

Re: provement

#16 2010-09-03 21:53:58

Re: provement

#17 2010-09-03 22:34:10

Re: provement

#18 2010-09-04 07:51:29

Re: provement

#19 2010-09-04 08:55:32

Re: provement

#20 2010-09-04 20:23:52

Re: provement

#21 2010-09-04 20:40:36

Re: provement

#22 2010-09-05 18:47:40

Re: provement

#23 2010-09-06 00:52:12

Re: provement

#24 2010-09-06 01:26:51

Re: provement

#25 2010-09-06 01:28:13

Re: provement

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