MCQ
If $(2^\text{n}+1)\text{x}=\pi,$ then $2^\text{n}\cos\text{x}\cos2\text{x}^2\text{x}\cos^\text{n-1}\text{x}=$
  • A
    $-1$
  • $1$
  • C
    $\frac{1}{2}$
  • D
    None of these

Answer

Correct option: B.
$1$
Given,
$(2^\text{n}+1)\text{x}=\pi$
$\Rightarrow2^\text{n}\text{x}+\text{x}=\pi$
$\Rightarrow2^\text{n}\text{x}=\pi-\text{x}$
$\Rightarrow\sin2^\text{n}\text{x}=\sin(\pi-\text{x})$
$\Rightarrow\sin2^\text{n}\text{x}=\sin\text{x}\ .....(1)$
$2^\text{n}\cos\text{x}\cos2\text{x}\cos2^2\text{x}\ ...\cos2^{\text{n}-1}\text{x}=2^\text{n}\times\frac{\sin2^\text{n}\text{x}}{2^\text{n}\sin\text{x}}$
$=\frac{\sin2^\text{n}\text{x}}{\sin\text{x}}$
$=\frac{\sin\text{x}}{\sin\text{x}}$ [From (1)]
$=1$

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