no experts are not allowed to answer this

Method to kafi simple ai.

solve he sreies of sin²âˆ‚ n cos²âˆ‚ differently after converting into double angle formulae thru C+iS method.

Will post soon................

first u must post the proof before quoting it simple
Prove that if the polynomial P(x)=a_0x^n+a_1x^{n-1}+\cdots\cdots+a_{n-1}x+a_nwith integral

coefficients has odd value for x=0 and x=1 , then the equation p(x)=0 can't have integral roots.

one more
the coefficient of x² in
\prod_{k=0}^{n}{\left(1+2^kx \right)}

i kno u wont accept this answer [3]

p(x)=a_{n}+a_{n-1}x+a_{n-2}x^{2}+a_{n-3}x^{3}+...+a_{o}x^{n}

p(0)=a_{n}=odd

p(2x)=a_{n}+a_{n-1}(2x)+a_{n-2}(2x)^{2}+a_{n-3}(2x)^{3}+...+a_{o}(2x)^{n}

thus p(2x) = a_n + even no. = odd + even no.
and odd + even can never be zero

hence no even solutions

now

p(2x+1)=a_{n}+a_{n-1}(2x+1)+a_{n-2}(2x+1)^{2}+a_{n-3}(2x+1)^{3}+...+a_{o}(2x+1)^{n}

now consider the term (2x+1)^k

(2x+1)^{k}= \sum_{r=0}^{k}{^{k}C_{r}}(2x)^{r}

(2x+1)^{k}= \sum_{r=0}^{k}{^{k}C_{r}}(2x)^{r}= ( 1 + ^{k}C_{2}(2x) +.....)

in this expansion only the first term i.e 1 is odd, rest all are even nos,

hence p(2x+1)=a_{n}+a_{n-1}(2x+1)+a_{n-2}(2x+1)^{2}+a_{n-3}(2x+1)^{3}+...+a_{o}(2x+1)^{n}

now take out 1 from each bracket of the type (2x+1)^k
\Rightarrow p(2x+1)=(a_{n}+a_{n-1}+a_{n-2}+...+a_{o})+ (a_{n-1}(2x) +a_{n-2}(4x^{2}+4x)+....)
=p(1)+ even \; no.

= odd + even

again wich can never be zero,

hence P(x) can never be zero for odd and even nos, i.e for integers