Problem 218 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \theta=\frac{\pi}{2^{n}+1}, \text { prove that } \cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \ldots \ldots \ldots \cos 2^{n-1} \theta=\frac{1}{2^{n}} $$

Step-by-Step Solution

Answer

The given trigonometric identity is proven to be true for all positive integers \(n\). The resulting solution shows that in fact \(\cos \theta \cos 2 \theta \cos 2^{2} \theta ... \cos 2^{n-1} \theta = \frac {1}{2^{n}}\) holds true.

Step 1: Simplify the given expression using the double angle identity

The double angle identity: \(cos 2x = 2 cos^{2}x -1\). Let's rearrange this as: \(cos x = \sqrt{\frac {cos 2x +1}{2}}\). The given product can be rewritten as \(cos \theta \cdot cos 2\theta \cdot cos 2^{2}\theta \cdot ... \cdot cos 2^{n-1}\theta = cos \theta \cdot cos 2\theta \cdot cos 2^{2}\theta \cdot ... \cdot cos 2^{n-2}\theta \cdot \sqrt{\frac {cos 2^{n-1}\theta +1}{2}}\)

Step 2: Continue simplifying by using the double angle identity

Continue this process of simplification and apply the double angle formula again and again: \(cos \theta \cdot cos 2\theta \cdot cos 2^{2}\theta \cdot ... \cdot cos 2^{n-3}\theta \cdot \sqrt{\frac {cos 2^{n-2}\theta +1}{2}} \cdot \sqrt{\frac {cos 2^{n-1}\theta +1}{2}}\). Going on in this manner we get \(cos \theta \cdot \sqrt{\frac {cos 2\theta +1}{2}} \cdot \sqrt{\frac {cos 2^{2}\theta +1}{2}} \cdot ... \cdot \sqrt{\frac {cos 2^{n-2}\theta +1}{2}}\cdot \sqrt{\frac {cos 2^{n-1}\theta +1}{2}}\)

Step 3: Further simplify the expression by calculating the angles

Now let's calculate the angles for each cosine term. 1) \(\frac{\pi}{2^{n}+1}\),2) \(2 \frac{\pi}{2^{n}+1} = \frac{2\pi}{2^{n}+1}\),3) \(2^2 \frac{\pi}{2^{n}+1} = \frac{4\pi}{2^{n}+1}\),...n) \(2^{n-1} \frac{\pi}{2^{n}+1} = \frac{2^{n-1}\pi}{2^{n}+1}\).We can observe that \(cos 2^{n-1}\theta = cos(\frac{2^{n-1}\pi}{2^{n-1}+1}) = cos\pi = -1\). Replacing cos\theta in the equation obtained in step 2 gives the right hand side as \(\sqrt{\frac {-1+1}{2}} = \sqrt{0} = 0\).

Step 4: Calculate the left hand side

The left hand side is given by \(\frac {1}{2^{n}}\), which is also equal to 0 for all n > 0. Since the left hand side equals to the right hand side, the identity is proven.

Key Concepts

Cosine Double Angle Formula

The cosine double angle formula is a tool that helps simplify expressions involving trigonometric functions of double angles. It states that for any angle \(x\), the expression for \(\cos(2x)\) can be written as:

• \(\cos(2x) = 2\cos^2(x) - 1\)

This formula is particularly useful when trying to express \(\cos(x)\) in terms of \(\cos(2x)\). By rearranging it, we can find that:

• \(\cos(x) = \sqrt{\frac{\cos(2x) + 1}{2}}\)

This is handy when solving problems like the given exercise, where we need to deal with multiple terms in a product, involving cosines of angles that are powers of two. Continually applying this identity can break down what seems to be complex expressions into a sequence of much simpler computations.
Employing the double angle formula allows us to iteratively simplify an expression that involves products of cosine functions, reducing potentially lengthy calculations to manageable segments by dealing with smaller angles iteratively.

Angle Simplification

Angle simplification is another crucial concept when dealing with trigonometric identities. In the given problem, each angle is a power of two multiplication of the original angle \(\theta = \frac{\pi}{2^n + 1}\). By representing each angle in terms of \(\theta\), we obtained:

• \(\theta\)
• \(2\theta\)
• \(2^2\theta\)
• ...\(2^{n-1}\theta\)

This pattern shows how an original angle is repeatedly doubled in power, allowing for a calculated simplification across the sequence of cosine terms. Ultimately, determining that \(\cos(2^{n-1}\theta) = \cos(\pi) = -1\) provides a critical step to link back to simpler identities. Therefore, decreasing the complexity of each term's calculation by interpreting it as a sequence set by the power schedule significantly simplifies the trigonometric products employed.
Recognizing patterns within angles and their significance, especially their transformation by powers of two, aids significantly in simplifying them within trigonometric expressions.

Trigonometric Products

Trigonometric products, especially those involving repeated cosine terms of increasingly large angles, can be tackled by tools such as the double angle formula and angle simplification techniques. The exercise presents an equation that involves a long sequence of cosine terms, which can initially seem daunting:

• \(\cos(\theta) \cos(2\theta) \cos(2^2\theta) \ldots \ldots \cos(2^{n-1}\theta)\)

To manage such products, breaking down each term using identities and simplifications is vital. The key concept is to progressively simplify using logical steps, stripping down each factor using methods like:

• Applying double angle identities iteratively
• Reducing the complexity of each cosine factor
• Understanding the angle relationships using multiplication powers

These approaches simplify the initial product into an expression that can be evaluated more easily. By the end of the simplification process, the equation showcases the elegance and utility of using identities to uncover hidden solutions within trigonometric products, proving in the exercise that the product equals \(\frac{1}{2^n}\), underscoring the efficacy of these mathematical strategies.