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Question

If $c o s invisible function application alpha plus 2 c o s invisible function application beta plus 3 c o s invisible function application gamma equals s i n invisible function application alpha plus 2 s i n invisible function application beta plus 3 s i n invisible function application gamma equals 0$ then $s i n invisible function application 3 alpha plus 8 s i n invisible function application 3 beta plus 27 s i n invisible function application 3 gamma equals$

$18 c o s (α + β + γ)$
$18 s i n (α + β + γ)$
$s i n (α + β + γ)$
$c o s (α + β + γ)$

The correct answer is: $18 s i n invisible function application left parenthesis alpha plus beta plus gamma right parenthesis$

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$1 comma alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma alpha subscript 3 horizontal ellipsis horizontal ellipsis end subscript alpha subscript n minus 1 end subscript$ are the n^th roots of unity and n is an even natural number, then $open parentheses 1 plus alpha subscript 1 end subscript close parentheses open parentheses 1 plus alpha subscript 2 end subscript close parentheses open parentheses 1 plus alpha subscript 3 end subscript close parentheses horizontal ellipsis horizontal ellipsis open parentheses 1 plus alpha subscript n minus 1 end subscript close parentheses equals$

$1 comma alpha subscript 1 end subscript comma alpha subscript 2 end subscript comma alpha subscript 3 horizontal ellipsis horizontal ellipsis end subscript alpha subscript n minus 1 end subscript$ are the n^th roots of unity and n is an even natural number, then $open parentheses 1 plus alpha subscript 1 end subscript close parentheses open parentheses 1 plus alpha subscript 2 end subscript close parentheses open parentheses 1 plus alpha subscript 3 end subscript close parentheses horizontal ellipsis horizontal ellipsis open parentheses 1 plus alpha subscript n minus 1 end subscript close parentheses equals$

Statement-I : If $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis equals negative 1$
Statement-II : $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 4 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis infinity equals negative 1$

Statement-I : If $x subscript n end subscript equals c o s invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses plus i s i n invisible function application open parentheses pi divided by 2 to the power of n end exponent close parentheses$ then $x subscript 1 end subscript times x subscript 2 end subscript times x subscript 3 end subscript horizontal ellipsis horizontal ellipsis equals negative 1$
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