Math Book Cambodia: It is given polynomial `P(x)=(xsina+cosa)^n` for natural number n. Finding the remaining when `P(x)` divide to `(x^2+1)`

Friday, November 27, 2020

It is given polynomial $P (x) = {(x sin a + cos a)}^{n}$ $P(x)=(xsina+cosa)^n$ for natural number n.

Finding the remaining when $P (x)$ $P(x)$ divide to $(x^{2} + 1)$ $(x^2+1)$

Solution

Let $R (x)$ $R(x)$ be the remainder of division between $P (x)$ $P(x)$ and $(x^{2} + 1)$ $(x^2+1)$

Therefore, the remainder

R (x) = x sin a + cos a

$R(x)=xsina+cosa$

If $n$ $n$ ≥ $2$ $2$ then $P (x) = (x^{2} + 1) Q (x) + R (x)$ $P(x)=(x^2+1)Q(x)+R(x)$

which

Q (x)

$Q(x)$ is the result and

R (x) = A x + B

$R(x)=Ax+B$

Then, ${(x sin a + cos a)}^{n} = (x^{2} + 1) Q (x) + A x + B$ $(xsina+cosa)^n=(x^2+1)Q(x)+Ax+B$

If $x = i$ $x=i$ then $cos n a + i sin n a = B + i A$ $cosna+isinna=B+iA$

Therefore, $A = sin (n a); B = cos (n a)$ $A=sin(na) ;B=cos(na)$

Hence, the remainder is $R (x) = sin (n a) x + c o n s (n a)$ $R(x)=sin(na)x+cons(na)$ .

Solution by: Thin Sokkean

Math Book Cambodia