Q193 / 270 · NDA 2018 (Phase 2) · Mathematics

If $\theta=\frac{\pi}{8}$, then what is the value of $(2 \cos \theta+1)^{10}(2 \cos 2 \theta-1)^{10}(2 \cos \theta-1)^{10}(2 \cos 4 \theta-1)^{10}$ ?

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    4

Solution

$[(2 \cos \theta+1)(2 \cos \theta-1)]^{10}[2 \cos 2 \theta-1]^{10}[2 \cos 4 \theta-1]^{10}$ when, $\theta=\frac{\pi}{8}$ (given) $\begin{aligned} & =\left(4 \cos ^2 \frac{\pi}{8}-1\right)^{10}\left(2 \times \frac{1}{\sqrt{2}}-1\right)^{10}(-1)^{10} \\ & =\left(4 \times \frac{2+\sqrt{2}}{4}-1\right)^{10}(\sqrt{2}-1)^{10} \\ & =[(\sqrt{2}+1)(\sqrt{2}-1)]^{10}=1^{10}=1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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