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The maximum value of $open parentheses cos invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cos invisible function application alpha subscript n end subscript close parentheses comma$ unser the restrictions $0 less or equal than alpha subscript 1 end subscript comma blank alpha subscript 2 end subscript comma horizontal ellipsis alpha subscript n end subscript less or equal than pi divided by 2$ and $open parentheses cot invisible function application alpha subscript 1 end subscript close parentheses open parentheses cot invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cot invisible function application alpha subscript n end subscript close parentheses equals 1$ is

$1 / 2^{n / 2}$
$1 / 2^{n}$
1/2 $n$
1

The correct answer is: $1 divided by 2 to the power of n divided by 2 end exponent$

We have given that
$open parentheses cot invisible function application alpha subscript 1 end subscript close parentheses open parentheses cot invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cot invisible function application alpha subscript n end subscript close parentheses equals 1$
$rightwards double arrow open parentheses cos invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cos invisible function application alpha subscript n end subscript close parentheses equals open parentheses sin invisible function application alpha subscript 1 end subscript close parentheses open parentheses sin invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis left parenthesis sin invisible function application alpha subscript n end subscript right parenthesis$
Let $y equals open parentheses cos invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cos invisible function application alpha subscript n end subscript close parentheses left parenthesis t o blank b e blank m a x i m u m right parenthesis$
Squaring both sides, we get
$y to the power of 2 end exponent equals open parentheses cos to the power of 2 end exponent invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos to the power of 2 end exponent invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis left parenthesis c o s subscript n end subscript superscript 2 end superscript right parenthesis$
$equals cos invisible function application alpha subscript 1 end subscript sin invisible function application alpha subscript 1 end subscript cos invisible function application alpha subscript 2 end subscript sin invisible function application alpha subscript 2 end subscript horizontal ellipsis cos invisible function application alpha subscript n end subscript sin invisible function application alpha subscript n end subscript$ [using Eq.(i)]
$equals fraction numerator 1 over denominator 2 to the power of n end exponent end fraction left square bracket sin invisible function application 2 alpha subscript 1 end subscript sin invisible function application 2 alpha subscript 2 end subscript horizontal ellipsis sin invisible function application 2 alpha subscript n end subscript right square bracket$
As $0 less or equal than 2 alpha subscript 1 end subscript comma blank 2 alpha subscript 2 end subscript comma horizontal ellipsis 2 alpha subscript 2 end subscript less or equal than pi$
$rightwards double arrow 0 less or equal than sin invisible function application 2 alpha subscript 1 end subscript comma sin invisible function application 2 alpha subscript 2 end subscript comma horizontal ellipsis comma sin invisible function application 2 alpha subscript n end subscript less or equal than 1$
$therefore y to the power of 2 end exponent less or equal than fraction numerator 1 over denominator 2 to the power of n end exponent end fraction 1 rightwards double arrow y less or equal than fraction numerator 1 over denominator 2 to the power of n divided by 2 end exponent end fraction$
Therefore, the maximum value of $y$ is $1 divided by 2 to the power of n divided by 2 end exponent$

Related Questions to study

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parallel

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parallel

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parallel

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Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

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parallel

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