Thread: m^2 and n^2 type problem for a perfect Genius!!

Given:-
m^2 + m’^2 2mm’cosΘ = 1
n^2 + n’^2 + 2nn’cosΘ = 1
mn + m’n’ + (mn’ + m’n) cosΘ = 0

To Prove:-
m^2 + n^2 = cosec^2Θ

Please help! Can anyone do this sum??


And one more important thing i want to say, is that I'm a class 10th student....
So please provide the easiest way round...

Hello, Lavlesh!

I haven't solved it yet, but I have some observations . . .

Given:
. . [1] . m² + m’² + 2mm’cosθ .= .1
. . [2] . n² + n’² + 2nn’cosθ .= .1
. . [3] .mn + m’n’ + (mn’ + m’n)cosθ .= .0

Prove: .m² + n² .= .csc²θ

Add [1], [2], and 2 times [3]:
. . . .[1] . m² + m’² + 2mm’cosθ .= .1
. . . .[2] . n² + n’² + 2nn’cosθ .= .1
. . 2x[3] .2mn + 2m’n’ + 2(mn’ + m’n)cosθ .= .0

Their sum is:
. . m² + 2mn + n² + m'² +2m'n' + n'² + 2(mm' + mn' + m'n + nn')cosθ .= .2

which simplifies to:
. . (m + n)² + (m' + n')² + 2(m + n)(m' + n')cosθ .= .2

Did that help? . . . Not a bit!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The given equations look suspiciously like the Law of Cosines.

I modified them:
. . [1] . m² + m’² - 2mm’cos(180°-θ) .= .1
. . [2] . n² + n’² - 2nn’cos(180°-θ) .= .1
. . [3] .mn + m’n’ - (mn’ + m’n)cos(180°-θ) .= .0

Adding [1], [2], and 2 times [3],

. . we get: .(m + n)² + (m' + n')² - 2(m + n)(m' + n')cos(180°-θ) .= .2


So I have a triangle . . .

Code:

      *
       *    *
        *         *   2
   m + n *              *
          *                   *
        θ  *                        *
      - - - *  *  *  *  *  *  *  *  *  *  *
                        m' + n'

But this too failed to inspire me.

The equations are not independtant.
Consider,
x=n=n'=m=m'
Then the first two equations give,
2x^2+2x^2cosθ=0
While the third gives,
2x^2+2x^2cosθ=1
This is an apparantet contradiction,
Implying the set of numbers is empty, (no wonder the great Soroban failed).

Good job done... Thanks...
But I still cannot understand the second method Mr. Soroban discussed...
But the first one given by him was pretty good!

But can this be a little more easier to grasp?? B'coz I'm in 10th class and I couldn't get that cos(180°-θ) and I've learnt only cos(90°-θ)
type methods....
Can you further explain the second method, you gave... Please....

Originally Posted by Lavlesh

Good job done... Thanks...
But I still cannot understand the second method Mr. Soroban discussed...
But the first one given by him was pretty good!

But can this be a little more easier to grasp?? B'coz I'm in 10th class and I couldn't get that cos(180°-θ) and I've learnt only cos(90°-θ)
type methods....
Can you further explain the second method, you gave... Please....

He was reffering to Law of Cosines. In order to use it you need a negative in front of the cosine. Thus, he used the identity,
cos(x)=-cos(180-x)

Originally Posted by ThePerfectHacker

The equations are not independtant.
Consider,
x=n=n'=m=m'
Then the first two equations give,
2x^2+2x^2cosθ=0
While the third gives,
2x^2+2x^2cosθ=1
This is an apparantet contradiction,
Implying the set of numbers is empty, (no wonder the great Soroban failed).

And can you explain what were you trying to do???

Originally Posted by Lavlesh

And can you explain what were you trying to do???

That there is no solution to your problem, it leads to a contradiction (mathematically correct to say it leads to an empty set of numbers). Thus, your problem is wrong and has no solutions.

EXAMPLE]
Given that,
x^2+y^2=2
x^4+y^4=0
Prove,
x^3+y^3=1

There is NO proof.

Look ar the second equation, the only solution is (0,0) because some of positives is zero. But that is false because look at the first.

Similarily what you are saying.
You are asking to prove something which is not true, which is why it is not provable.