What is the value of ≤ft(5π/17)+≤ft(7π/17)+2≤ft(11π/17)≤ft(π/17) ?

4 cos≤ft(6π/17) cos≤ft(π/17)

4 cos≤ft(11π/17) cos≤ft(π/17)

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UPSC NDA 2 2022 Mathematics Solved Paper

Question : 47 of 120

Marks: +1, -0

\cos\left(\frac{5\pi}{17}\right)+\cos\left(\frac{7\pi}{17}\right)+2\cos\left(\frac{11\pi}{17}\right)\cos\left(\frac{\pi}{17}\right)

Solution:

Concept:

Use the sum-to-product identity:

\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)

Explanation:

Given expression:

\cos\left(\frac{5\pi}{17}\right) + \cos\left(\frac{7\pi}{17}\right) + 2\cos\left(\frac{11\pi}{17}\right) \cos\left(\frac{\pi}{17}\right)

Apply the identity to the first two terms:

2 \cos\left(\frac{6\pi}{17}\right) \cos\left(-\frac{\pi}{17}\right) + 2\cos\left(\frac{11\pi}{17}\right) \cos\left(\frac{\pi}{17}\right)

Since

\cos(-\theta) = \cos\theta

, factor out

2\cos\left(\frac{\pi}{17}\right)

2\cos\left(\frac{\pi}{17}\right)\left[\cos\left(\frac{6\pi}{17}\right) + \cos\left(\frac{11\pi}{17}\right)\right]

Apply the identity again inside the bracket:

2\cos\left(\frac{\pi}{17}\right) \cdot 2 \cos\left(\frac{17\pi}{34}\right) \cos\left(-\frac{5\pi}{34}\right)

4\cos\left(\frac{\pi}{17}\right) \cos\left(\frac{\pi}{2}\right) \cos\left(\frac{5\pi}{34}\right)

\cos\left(\frac{\pi}{2}\right) = 0

, so the entire product is

0

Answer:

0

(Option A).

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