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Maths-

General

Easy

Question

If $theta equals fraction numerator pi over denominator 2 to the power of n plus 1 end fraction$ then $cos space theta cos space 2 theta cos space 2 squared theta horizontal ellipsis cos space 2 to the power of n minus 1 end exponent theta$ is equal to

$\frac{1}{2^{n}}$
$\cos θ$
2
$2^{n}$

hint Hint:

$2 cos theta sin theta equals sin 2 theta$

The correct answer is: $1 over 2 to the power of n$

$theta equals fraction numerator pi over denominator 2 to the power of n plus 1 end fraction theta open parentheses 2 to the power of n plus 1 close parentheses equals straight pi 2 to the power of straight n straight theta plus straight theta equals straight pi 2 to the power of straight n straight theta equals straight pi minus straight theta space space space.... open parentheses 1 close parentheses cos space θcos space 2 θcos space 2 squared straight theta horizontal ellipsis cos space 2 to the power of straight n minus 1 end exponent straight theta equals fraction numerator 2 sinθ over denominator 2 sinθ end fraction open parentheses cos space θcos space 2 θcos space 2 squared straight theta horizontal ellipsis cos space 2 to the power of straight n minus 1 end exponent straight theta close parentheses equals fraction numerator 1 over denominator 2 sinθ end fraction open parentheses 2 sinθcos space θcos space 2 θcos space 2 squared straight theta horizontal ellipsis cos space 2 to the power of straight n minus 1 end exponent straight theta close parentheses open parentheses because 2 sinθcosθ equals sin 2 straight theta close parentheses equals fraction numerator 1 over denominator 2 sinθ end fraction open parentheses sin 2 θcos space 2 θcos space 2 squared straight theta horizontal ellipsis cos space 2 to the power of straight n minus 1 end exponent straight theta close parentheses equals fraction numerator 1 over denominator 2 squared sinθ end fraction open parentheses sin 2 squared θcos space 2 squared straight theta horizontal ellipsis cos space 2 to the power of straight n minus 1 end exponent straight theta close parentheses equals fraction numerator 1 over denominator 2 to the power of straight n minus 1 end exponent sinθ end fraction open parentheses sin 2 to the power of straight n minus 1 end exponent θcos space 2 to the power of straight n minus 1 end exponent straight theta close parentheses equals fraction numerator 1 over denominator 2 to the power of straight n sinθ end fraction open parentheses sin 2 to the power of straight n straight theta close parentheses equals fraction numerator sin 2 to the power of straight n straight theta over denominator 2 to the power of straight n sinθ end fraction From space equation space 1 comma equals fraction numerator sin open parentheses straight pi minus straight theta close parentheses over denominator 2 to the power of straight n sinθ end fraction open parentheses because sin open parentheses straight pi minus straight theta close parentheses equals sinθ close parentheses equals fraction numerator sinθ over denominator 2 to the power of straight n sinθ end fraction equals 1 over 2 to the power of straight n$

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