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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2025 · Mathematics · Trigonometric Ratios & Identities

Consider the following for the two (02) items that follow: Let 2 α...

Consider the following for the two (02) items that follow:
Let 2sin⁡α+cos⁡α=22\sin\alpha + \cos\alpha = 2, where 0<α<90°0 < \alpha < 90°.

What is 2sin⁡2α+cos⁡2α2\sin 2\alpha + \cos 2\alpha equal to?

Options

  1. A1110\dfrac{11}{10}
  2. B115\dfrac{11}{5}
  3. C125\dfrac{12}{5}
  4. D135\dfrac{13}{5}

Correct answer

B. 115\dfrac{11}{5}

Step-by-step solution

Given 2sin⁡α+cos⁡α=22\sin\alpha + \cos\alpha = 2

cos⁡α=2(1−sin⁡α)\cos\alpha = 2(1 - \sin\alpha)

Squaring both sides:

cos⁡2α=4(1−sin⁡α)2\cos^2\alpha = 4(1 - \sin\alpha)^2

1−sin⁡2α=4(1−2sin⁡α+sin⁡2α)1 - \sin^2\alpha = 4(1 - 2\sin\alpha + \sin^2\alpha)

5sin⁡2α−8sin⁡α+3=05\sin^2\alpha - 8\sin\alpha + 3 = 0

(5sin⁡α−3)(sin⁡α−1)=0(5\sin\alpha - 3)(\sin\alpha - 1) = 0

Since 0 < \alpha < 90^{}^{\circ}, sin⁡α≠1\sin\alpha \neq 1. Thus, sin⁡α=35\sin\alpha = \dfrac{3}{5}.

This gives cos⁡α=1−(35)2=45\cos\alpha = \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = \dfrac{4}{5}.

Now, sin⁡2α=2sin⁡αcos⁡α=2(35)(45)=2425\sin 2\alpha = 2\sin\alpha\cos\alpha = 2\left(\dfrac{3}{5}\right)\left(\dfrac{4}{5}\right) = \dfrac{24}{25}

cos⁡2α=cos⁡2α−sin⁡2α=(45)2−(35)2=725\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(\dfrac{4}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2 = \dfrac{7}{25}

Therefore, 2sin⁡2α+cos⁡2α=2(2425)+725=48+725=5525=1152\sin 2\alpha + \cos 2\alpha = 2\left(\dfrac{24}{25}\right) + \dfrac{7}{25} = \dfrac{48 + 7}{25} = \dfrac{55}{25} = \dfrac{11}{5}.

Answer: 115\dfrac{11}{5}
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