if x2- 2cos∂ x +1 = 0 and y2 - 2 cos∞ y + 1 = 0 then 2cos(∂+∞) is equal to
49is it x2 - (2cos∂)x + 1 = 0
and
y2 - (2cos∞)y + 1 = 0
or,
x2 - 2cos(∂x) + 1 = 0
and
y2 - 2cos(∞y) + 1 = 0
???
49for the first set of equations i have written,
cos ∂ = x2+12x so, sin ∂ = √(1-(x2+1)2(2x)2)
cos ∞ = y2+12y so, sin ∞ = √(1-(y2+1)2(2y)2)
cos(∂+∞)=cos∂cos∞-sin∂sin∞
62from where subhomoy has left, RHS in the expression of cos in #3,
RHS>=1
So cos a=1 and cos b =1
sin a =0, cos a =0
So we have 2cos(a+b)=2
49exactly that is what i was thinking and so left the question incomplete!!
actually was sort of confused about sine function which was seemingly imaginary ;)
341x and y need not be real. x = ei∂ does the job.
The "right" answer would be
\frac{1}{2} \left[ \left(x + \frac{1}{x}\right) \left(y + \frac{1}{y}\right) + \left|x - \frac{1}{x} \right| \left| y - \frac{1}{y} \right| \right]
341which means they have naively chosen
\frac{1}{2} \left[ \left(x + \frac{1}{x} \right)\left(y+\frac{1}{y} \right) + \left(x - \frac{1}{x} \right)\left(y-\frac{1}{y} \right) \right]
BTW, my previous post needs an edit. By mistake, between the two expressions there should be a ±. and no mod for second expression.
