Theta is just a random value and c is a random constant just find n
Can you also elaborate on the RHS using some kind of summation notation? It is not clear what fills the "..." given that there are cosines at the start but not the end.Self made question... goodluck:
 +ncos((n-2))\theta+...+^nC_{\frac{n}{2}} (n-n)\theta +c )
You can find n without using in terms of c and theta
i can answer your question
Oh shit it should have cos(n-n)theta at the end my bad
this does not make sense to me, whatever your answer is, say n=2, the equation can't be true for random c.
Well the answers n=0i can answer your question
but im extremely confused about how theta and c are random
fixed versionSelf made question... goodluck
 +ncos((n-2))\theta+...+^nC_{\frac{n}{2}}cos(n-n)\theta +c )
Using standard complex numbers tricks (raise a complex number on the unit circle plus it's reciprocal to the power 2n in two different ways) and exploiting that cos is even we can write the RHS as:
No, you certainly can't. Otherwise suppose the equation is true for a given n and for any c and any theta.
you do use:Using standard complex numbers tricks (raise a complex number on the unit circle plus it's reciprocal to the power 2n in two different ways) and exploiting that cos is even we can write the RHS as:
.
Then.
Then for any n such that the left hand side is positive and does not exceed 1, we take the 2 (or 1) real (2n)-th roots, and apply the general formula to solve cos(x)=a.
This is the complete set of solutions for the problem as posed.
Note that there are an ODD number of terms in (z+1/z)^(2n).you do use:
No, it is just a relabelling so we don't have to always pay attention to n's parity. This is different to squaring both sides of your equation. (Which is generally invalid as a solution method.)
