Question 1095761
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if 2costheta=x=1/x, 2cosphi=y+1/y,2coslamida=z+1/z,then prove that 2cos(theta+phi+lamida)=xyz+1/(xyz).
plz send me ans as early as possible.
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Obviously,   the post contains the error  (the typo ?)  in this part   "2costheta = x=1/x",  which must be read as  "2costheta = x+(1/x)".


With this editing, the solution is as follows.



<pre>
1.  If  2*cos(theta) = x + (1/x),  it implies that 


    EITHER  x = 1   and  then  2cos(theta) = 1 + (1/1) = 2  ====>  cos(theta) = 1  ====>  theta = 0,

    OR      x = -1  and  then  2cos(theta) = -1 + (1/(-1)) = -2  ====>  cos(theta) = -1  ====>  theta = {{{pi}}}.


   It is simply because  x + (1/x) >= 2  if x is positive,  or

                         x + (1/x) <= -2  if x is negative,

   which is  <U>VERY WELL KNOWN fact</U>.



2.  Similarly, from the given data 

    cos(phi) = +/-1      and  phi    = 0     or  phi    =  {{{pi}}}.

    cos(lambda) = +/-1   and  lambda = 0     or  lambda =  {{{pi}}}.



3.  Now, if  theta=0,  phi=0  and  lambda=0
         (equivalently, all three cosines are equal to 1)

    then  2cos(theta+phi+lamida)=xyz+1/(xyz)  is, obviously, true.



    Also, if  theta={{{pi}}},  phi={{{pi}}}  and  lambda={{{pi}}}
         (equivalently, all three cosines are equal to -1)

    then  2cos(theta+phi+lamida)=xyz+1/(xyz)  is, obviously, true, again.


    Finally, if any other combination of possible angles is chosen, 
    it will lead to the same equality, as it is easy to check, having very restricted number of combinations.
</pre>

SOLVED.



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<pre>
The key idea of the solution above is this fact:


    if x is the positive real number, then  {{{x + 1/x}}} >=2,  
    
    and the equality takes place if and only if  x = 1.


<U>Proof</U>


{{{x + 1/x}}} = {{{(sqrt(x) - 1/sqrt(x))^2 + 2}}} is greater than or equal to 2.

The equality takes place if and only if  {{{(sqrt(x) - 1/sqrt(x))^2}}} = 0,  which is equivalent to 

{{{sqrt(x) - 1/sqrt(x)}}} = 0  <=== equivalent to  ===>  {{{sqrt(x)}}} = {{{1/sqrt(x)}}}    <=== equivalent to  ===>  x = 1  (under the assumption that x is positive).


The proof is completed.
</pre>


<pre>
    <U>Notice</U>.  If you have questions or need more help, you can send a message to me through the "Thank you" form note.

             In this case PLEASE do not forget to refer to the problem ID number, which is  1095761,  
             in order for I could identify the problem.
</pre>