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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.
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 =
.
It is simply because x + (1/x) >= 2 if x is positive, or
x + (1/x) <= -2 if x is negative,
which is VERY WELL KNOWN fact.
2. Similarly, from the given data
cos(phi) = +/-1 and phi = 0 or phi =
.
cos(lambda) = +/-1 and lambda = 0 or lambda =
.
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=
, phi=
and lambda=
(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.
SOLVED.
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The key idea of the solution above is this fact:
if x is the positive real number, then
>=2,
and the equality takes place if and only if x = 1.
Proof
=
is greater than or equal to 2.
The equality takes place if and only if
= 0, which is equivalent to
= 0 <=== equivalent to ===>
=
<=== equivalent to ===> x = 1 (under the assumption that x is positive).
The proof is completed.
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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.