Problem 1 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\cos\) a \(\cos 2 a \cos 3 a \ldots . \cos 999 a\), where \(a=\frac{2 \pi}{1999}\), is (A) \(\frac{1}{2^{99}}\) (B) \(\frac{1}{2^{999}}\) (C) \(\frac{1}{2^{9999}}\) (D) \(\frac{1}{2^{1999}}\)

Step-by-Step Solution

Answer

The value is \( \frac{1}{2^{999}} \), which is option (B).

Step 1: Understanding the Product

We need to evaluate the product \( \cos a \cos 2a \cos 3a \ldots \cos 999a \), where \( a = \frac{2\pi}{1999} \). This expression involves the cosine of integer multiples of \( a \).

Step 2: Identifying Symmetry in Cosine Function

The cosine function is periodic with period \( 2\pi \). Additionally, it possesses symmetry properties such as \( \cos(\pi - x) = -\cos(x) \). This symmetry will be useful in simplifying the product.

Step 3: Utilizing the Symmetry

The given range of angles from \( a \) to \( 999a \) covers half of a complete circle. Since \( a = \frac{2\pi}{1999} \), multiplying \( a \) by \( 999 \) gives us \( \frac{999 \times 2\pi}{1999} \), which is less than \( \pi \).

Step 4: Apply the Formula for Symmetric Products

The angles \( a, 2a, ..., 999a \) are equidistant points along the arc \( [0, \pi] \). The product \( \prod_{k=1}^{999} \cos \left( \frac{2k\pi}{1999} \right) \) can be evaluated using known trigonometric identities for products. We have \( \prod_{k=1}^{n} \cos\left( \frac{(2k-1)\pi}{2n} \right) = \frac{1}{2^{n-1}} \) when angles are symmetrically distributed in \( [0, \pi] \).

Step 5: Calculation for the given problem

In our case, \( n = 999 \), so the product is \( \prod_{k=1}^{999} \cos \left( \frac{2\pi k}{1999} \right) = \frac{1}{2^{999}} \).

Step 6: Conclusion

Therefore, the given product evaluates to \( \frac{1}{2^{999}} \), which corresponds to option (B).

Key Concepts

Cosine Function

The cosine function is a fundamental trigonometric function denoted as \( \cos \). It relates an angle in a right-angled triangle to the ratio of the length of the adjacent side over the hypotenuse. The cosine function is periodic, which means it repeats its values over intervals. Specifically, its period is \( 2\pi \), and it is symmetric about the y-axis.\
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Understanding the cosine function is essential when dealing with problems involving angles, especially when multiple angles are multiples of a single angle, like in the given exercise. The key properties include periodicity and symmetry, which greatly simplify the evaluation of expressions involving many cosine terms.

Trigonometric Products

Finding products of trigonometric functions like the cosine function can involve complex identities that help simplify the problem. When dealing with a series of cosine terms like \( \cos a \cos 2a \cos 3a \ldots \cos 999a \), it's crucial to recognize patterns and apply known identities to simplify the calculation.\
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One such identity useful here is the product-to-sum identities. Moreover, specific formulas exist for symmetric arrangements of angles, like the one used in the original solution. This formula allows the simplification of the product by exploiting symmetry, reducing repetitive calculations and revealing the underlying reduction structure.

Symmetry in Trigonometry

In trigonometry, symmetry plays a vital role in simplifying expressions and understanding function behaviors. Cosine, in particular, exhibits useful symmetry properties:

• \( \cos(\pi - x) = -\cos(x) \)
• \( \cos(x) \) is an even function, meaning \( \cos(-x) = \cos(x) \)

These symmetric properties allow us to predict the values cosine will take over specific intervals, like when angles are distributed symmetrically over \([0, \pi]\). Such was the approach applied in solving the given problem.\
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By knowing the distribution of angles across a half-circle, and applying the symmetry of the cosine function, we can harness the fact that half of these cosines will negate the others across that interval, leading to significant simplification. Thus, symmetry is not just an abstract concept; it provides practical routes to solving otherwise complex trigonometric products efficiently.