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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

If α+ α=p , then what is ^2(2 α) equal to?

If sin⁡α+cos⁡α=p\sin \alpha+\cos \alpha=p, then what is cos⁡2(2α)\cos ^2(2 \alpha) equal to?

Options

  1. Ap2\mathrm{p}^2
  2. Bp2−1\mathrm{p}^2-1
  3. Cp2(2−p2)\mathrm{p}^2 \left(2- \mathrm{p}^2\right)
  4. Dp2+1\mathrm{p}^2+1

Correct answer

C. p2(2−p2)\mathrm{p}^2 \left(2- \mathrm{p}^2\right)

Step-by-step solution

sin⁡α+cos⁡α=p\sin \alpha+\cos \alpha=p⇒(sin⁡α+cos⁡α)2=p2⇒sin⁡2α+cos⁡2α+2sin⁡αcos⁡α=p2⇒1+sin⁡2α=p2⇒sin⁡2α=p2−1cos⁡22α=1−sin⁡22α=1−(p2−1)2=1−(p4+1−2p2)=−p4+2p2=p2(2−p2)\begin{aligned} & \Rightarrow(\sin \alpha+\cos \alpha)^2=\mathrm{p}^2 \\ & \Rightarrow \sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha \cos \alpha=\mathrm{p}^2 \\ & \Rightarrow 1+\sin 2 \alpha=\mathrm{p}^2 \\ & \Rightarrow \sin 2 \alpha=\mathrm{p}^2-1 \\ & \cos ^2 2 \alpha=1-\sin ^2 2 \alpha=1-\left(\mathrm{p}^2-1\right)^2 \\ & =1-\left(\mathrm{p}^4+1-2 \mathrm{p}^2\right)=-\mathrm{p}^4+2 \mathrm{p}^2 \\ & =\mathrm{p}^2\left(2-\mathrm{p}^2\right)\end{aligned}$$
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