Prove that if $c o s α + s i n α$ is a solution of the equation
$x^{n} + P_{1} X^{n - 1} + p_{2} X^{n - 2} + . . . . + P_{n} = 0$
then $P_{1} s i n α + P_{2} s i n 2 α + . . . . + P_{n} s i n n α = 0$
( P_1 , P_2 , ......, P_n are real)

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Solution

Since cos $α$ + i sin $α$ is the root of the given equation, we have
$(c o s α + i s i n α)^{n} + p_{1} (c o s α + i s i n α)^{n - 1}$
$+ p_{2} (c o s α + i s i n α)^{n - 2} + . . . p_{n} = 0$
$(c o s α + i s i n α)^{n} [1 + p_{1} e^{i α} + p_{2} e^{- 2 i α} + p_{3} e^{- 3 i α} + . . . . . + p_{n} e^{- 2 i α} = 0$
Now cos $α$ + i sin $α$ $\neq$ 0 for any $α$ , we can cancel the factor $(c o s α + i s i n α)^{n}$ in the equation (1). hence (1) can be written as
$1 + p_{1} (c o s α - i s i n α) + p_{2} (c o s 2 α - i s i n 2 α) + p_{3} (c o s 3 α - i s i n 3 α) + . . . . . . + p_{n} (c o s n α - i s i n n α) = 0$
Equating imaginary part to 0 in (2), we get
$p_{1} s i n α + p_{2} s i n 2 α + p_{3} s i n 3 α + . . . . . p_{n} s i n n α = 0.$
Equating real parts in (2), we shall obtain
$1 + p_{1} c o s α + p_{2} c o s 2 α + . . . . . p_{n} c o s n α = 0$

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