Test Index

UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 73-74

Consider the following for the next items that follow:

Let

p=cos(\frac{\pi }{5})cos(\frac{2\pi }{5})

and

q=cos(\frac{4\pi }{5})cos(\frac{8\pi }{5})

Question : 74 of 120

Marks: +1, -0

What is the value of pq?

Solution:

Concept:

Use complementary angle identities:

\cos(\pi - A) = -\cos A

and

\cos(2\pi - A) = \cos A

Also,

\cos 36^\circ = \frac{\sqrt{5}+1}{4}

and

\cos 72^\circ = \frac{\sqrt{5}-1}{4}

Explanation:

We have

p = \cos\frac{\pi}{5} \cos\frac{2\pi}{5}

and

q = \cos\frac{4\pi}{5} \cos\frac{8\pi}{5}

Simplify

q

\cos\frac{4\pi}{5} = \cos(\pi - \frac{\pi}{5}) = -\cos\frac{\pi}{5}

\cos\frac{8\pi}{5} = \cos(2\pi - \frac{2\pi}{5}) = \cos\frac{2\pi}{5}

Thus

q = (-\cos\frac{\pi}{5}) \cdot \cos\frac{2\pi}{5} = -\cos\frac{\pi}{5} \cos\frac{2\pi}{5}

Then

pq = (\cos\frac{\pi}{5} \cos\frac{2\pi}{5}) \times (-\cos\frac{\pi}{5} \cos\frac{2\pi}{5}) = -\cos^2\frac{\pi}{5} \cos^2\frac{2\pi}{5}

Substitute the known values:

\cos\frac{\pi}{5} = \cos 36^\circ = \frac{\sqrt{5}+1}{4}

and

\cos\frac{2\pi}{5} = \cos 72^\circ = \frac{\sqrt{5}-1}{4}

pq = -\left(\frac{\sqrt{5}+1}{4}\right)^2 \times \left(\frac{\sqrt{5}-1}{4}\right)^2

= -\left[\frac{(\sqrt{5}+1)(\sqrt{5}-1)}{16}\right]^2 = -\left(\frac{5-1}{16}\right)^2 = -\left(\frac{4}{16}\right)^2 = -\left(\frac{1}{4}\right)^2 = -\frac{1}{16}

Answer:

-\frac{1}{16}

Go to Question:

Prev Question

Next Question