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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If θ=(π)/(8) , then what is the value of (2 θ+1)^ 10 (2 2 θ-1)^ 10...

If θ=π8\theta=\frac{\pi}{8}, then what is the value of (2cos⁡θ+1)10(2cos⁡2θ−1)10(2cos⁡θ−1)10(2cos⁡4θ−1)10(2 \cos \theta+1)^{10}(2 \cos 2 \theta-1)^{10}(2 \cos \theta-1)^{10}(2 \cos 4 \theta-1)^{10} ?

Options

  1. A0
  2. B1
  3. C2
  4. D4

Correct answer

B. 1

Step-by-step solution

[(2cos⁡θ+1)(2cos⁡θ−1)]10[2cos⁡2θ−1]10[2cos⁡4θ−1]10[(2 \cos \theta+1)(2 \cos \theta-1)]^{10}[2 \cos 2 \theta-1]^{10}[2 \cos 4 \theta-1]^{10} when, θ=π8\theta=\frac{\pi}{8} (given) =(4cos⁡2π8−1)10(2×12−1)10(−1)10=(4×2+24−1)10(2−1)10=[(2+1)(2−1)]10=110=1\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}
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