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

  1. 1 over 2 to the power of n
  2. cos space theta
  3. 2
  4. 2 to the power of n

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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