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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 43 of 120

Marks: +1, -0

If

\theta =\frac{\pi }{8},

Then what is the value of

(2\cos \theta + 1)^{10} (2\cos 2\theta - 1)^{10} (2\cos \theta - 1)^{10} (2\cos 4\theta - 1)^{10}

?

0
1
2
4

Solution:

Concept:

Use algebraic grouping and trigonometric values to simplify the product.

Explanation:

First, group the terms

(2\cos\theta + 1)

and

(2\cos\theta - 1)

.

Their product is

(2\cos\theta)^2 - 1^2 = 4\cos^2\theta - 1

.

The expression becomes

(4\cos^2\theta - 1)^{10} (2\cos2\theta - 1)^{10} (2\cos4\theta - 1)^{10}

.

Now substitute

\theta = \frac{\pi}{8}

.

Then

2\theta = \frac{\pi}{4}

and

4\theta = \frac{\pi}{2}

.

Use known values:

\cos^2\frac{\pi}{8} = \frac{2+\sqrt{2}}{4}

,

\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}

,

\cos\frac{\pi}{2} = 0

.

Compute

4\cos^2\frac{\pi}{8} - 1 = 4\cdot\frac{2+\sqrt{2}}{4} - 1 = 2+\sqrt{2} - 1 = 1+\sqrt{2}

.

Compute

2\cos\frac{\pi}{4} - 1 = 2\cdot\frac{1}{\sqrt{2}} - 1 = \sqrt{2} - 1

.

Compute

2\cos\frac{\pi}{2} - 1 = 0 - 1 = -1

, and

(-1)^{10} = 1

.

So the product is

(1+\sqrt{2})^{10} (\sqrt{2}-1)^{10} \cdot 1

.

Combine:

[(1+\sqrt{2})(\sqrt{2}-1)]^{10} = [(2 - 1)]^{10} = 1^{10} = 1

.

Answer:

1

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