Trigonometric Series

Sum of SINES of Angles in Arithmetic Progression

Let \( S_n \) represent the sum of sine functions whose angles are in an arithmetic progression, such that:

\[ S_n = \sin(a) + \sin(a + d) + \sin(a + 2d) + \ldots + \sin[a + (n - 1)d] \]

where:

• \( a \) is the first term of the arithmetic progression,
• \( d \) is the common difference between the terms,
• \( n \) is the number of terms.

The sum \( S_n \) of the sine functions of \( n \) angles in arithmetic progression can be expressed as:

\[ S_n = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \sin\left(a + \frac{(n - 1)d}{2}\right) \]

Proof:

1. Multiply each term in the series by \( 2\sin(\frac{d}{2}) \):

   \[ 2\sin\left(\frac{d}{2}\right)S = 2\sin\left(\frac{d}{2}\right)\sin(a) + 2\sin\left(\frac{d}{2}\right)\sin(a + d) + \ldots + 2\sin\left(\frac{d}{2}\right)\sin[a + (n - 1)d] \]

2. Apply the product-to-sum formula to each term:

   \[ \sin(x)\sin(y) = \frac{1}{2}[\cos(x - y) - \cos(x + y)] \]

   Hence, we transform each term:

   \[ 2\sin\left(\frac{d}{2}\right)\sin(a + kd) = \cos\left(a + kd - \frac{d}{2}\right) - \cos\left(a + kd + \frac{d}{2}\right) \]

3. Now, the series becomes a telescoping series:

   \[ 2\sin\left(\frac{d}{2}\right)S = [\cos\left(a - \frac{d}{2}\right) - \cos\left(a + \frac{d}{2}\right)] + [\cos\left(a + \frac{d}{2}\right) - \cos\left(a + \frac{3d}{2}\right)] + \ldots + [\cos\left(a + (n - 2)\frac{d}{2}\right) - \cos\left(a + n\frac{d}{2}\right)] \]

4. In the above expression, each \( \cos \) term cancels the subsequent term except for the first negative term and the last positive term:

   \[ \begin{aligned} & \cos\left(a - \frac{d}{2}\right) & - \color{red}\cancel{\cos\left(a + \frac{d}{2}\right)} \\ & \color{red}\cancel{\cos\left(a + \frac{d}{2}\right)} & - \color{red}\cancel{\cos\left(a + \frac{3d}{2}\right)} \\ & \color{red}\cancel{\cos\left(a + \frac{3d}{2}\right)} & - \color{red}\cancel{\cos\left(a + \frac{5d}{2}\right)} \\ & \vdots & \vdots \\ & \color{red}\cancel{\cos\left(a + (n-3)\frac{d}{2}\right)} & - \color{red}\cancel{\cos\left(a + (n-2)\frac{d}{2}\right)} \\ & \color{red}\cancel{\cos\left(a + (n-2)\frac{d}{2}\right)} & - \cos\left(a + n\frac{d}{2}\right) \\ \end{aligned} \]

   Here, each term that is struck through is canceled by a corresponding term on a subsequent line. After the cancellation, the first term on the first line and the last term on the last line remain, leading to the simplified result:

   \[ 2\sin\left(\frac{d}{2}\right)S = \cos\left(a - \frac{d}{2}\right) - \cos\left(a + (n - 1)\frac{d}{2}\right) \]

5. The final step is to express the difference of cosines back into a product of sines, using the sum-to-product identities:

   \[ \cos(x) - \cos(y) = -2\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right) \]

   So we have:

   \[ 2\sin\left(\frac{d}{2}\right)S = -2\sin\left(\frac{a - \frac{d}{2} + a + (n - 1)\frac{d}{2}}{2}\right)\sin\left(\frac{a - \frac{d}{2} - (a + (n - 1)\frac{d}{2})}{2}\right) \]
   \[ 2\sin\left(\frac{d}{2}\right)S = -2\sin\left(a + \frac{(n - 1)d}{2}\right)\sin\left(-\frac{nd}{2}\right) \]

6. Simplify and solve for \( S \):

   \[ S = -\frac{\sin\left(a + \frac{(n - 1)d}{2}\right)\sin\left(-\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]

   Since \( \sin(-x) = -\sin(x) \), we get:

   \[ S = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)}\sin\left(a + \frac{(n - 1)d}{2}\right) \]

Example

Problem: Prove that the sum of the series \( S = \sin(\theta) + \sin(3\theta) + \sin(5\theta) + \ldots + \sin((2n-1)\theta) \) is equal to \( \frac{\sin^2(n\theta)}{\sin(\theta)} \).

Solution: We can use the formula for the sum of a sine series where the angles are in arithmetic progression:

\[ S = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \sin\left(a + \frac{(n - 1)d}{2}\right) \]

In the given series, the first term \( a \) is \( \theta \), the common difference \( d \) is \( 2\theta \), and there are \( n \) terms because the series goes from \( \sin(\theta) \) to \( \sin((2n-1)\theta) \).

Let's identify the components for the formula: 1. The term \( \frac{nd}{2} \) becomes \( \frac{n \cdot 2\theta}{2} = n\theta \). 2. The term \( \frac{d}{2} \) becomes \( \frac{2\theta}{2} = \theta \). 3. The term \( a + \frac{(n - 1)d}{2} \) becomes \( \theta + \frac{(n - 1) \cdot 2\theta}{2} = \theta + (n - 1)\theta = n\theta \).

Now the sum \( S \) is:

\[ S = \frac{\sin(n\theta)}{\sin(\theta)} \cdot \sin(n\theta) \]

This simplifies to:

\[ S = \frac{\sin^2(n\theta)}{\sin(\theta)} \]

Thus, we have proved that \( S = \sin(\theta) + \sin(3\theta) + \sin(5\theta) + \ldots + \sin((2n-1)\theta) \) is indeed equal to \( \frac{\sin^2(n\theta)}{\sin(\theta)} \).

Sum of COSINES of Angles in Arithmetic Progression

The sum of a cosine series with angles in an arithmetic progression is given by:

\[ S = \cos(a) + \cos(a + d) + \cos(a + 2d) + \ldots + \cos[a + (n - 1)d] \]

This sum can be expressed as:

\[ S = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \cos\left(a + \frac{(n - 1)d}{2}\right) \]

Proof:

Given the series:

\[ S = \cos(a) + \cos(a + d) + \cos(a + 2d) + \ldots + \cos[a + (n - 1)d] \]

We want to show that:

\[ S = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \cos\left(a + \frac{(n - 1)d}{2}\right) \]

Step 1: Multiply each term by \( 2\sin\left(\frac{d}{2}\right) \)

\[ 2\sin\left(\frac{d}{2}\right)S = 2\sin\left(\frac{d}{2}\right)\cos(a) + 2\sin\left(\frac{d}{2}\right)\cos(a + d) + \ldots + 2\sin\left(\frac{d}{2}\right)\cos[a + (n - 1)d] \]

Step 2: Apply the product-to-sum formula

The product-to-sum formula is \( 2\cos A \sin B = \sin(A + B) - \sin(A - B) \). Apply this to each term:

\[ 2\sin\left(\frac{d}{2}\right)\cos(a + kd) = \sin\left(a + kd + \frac{d}{2}\right) - \sin\left(a + kd - \frac{d}{2}\right) \]

where \(k=0, 1, 2, ..., n-1\)

Step 3: Form a telescoping series

After applying the product-to-sum formula, we have a series where each term cancels the next one, except for the first and the last terms:

\[ 2\sin\left(\frac{d}{2}\right)S = \left[\sin\left(a + \frac{d}{2}\right) - \sin\left(a - \frac{d}{2}\right)\right] + \left[\sin\left(a + \frac{3d}{2}\right) - \sin\left(a + \frac{d}{2}\right)\right] + \ldots + \left[\sin\left(a + (n - \frac{1}{2})d\right) - \sin\left(a + (n - \frac{3}{2})d\right)\right] \]

Step 4: Apply the telescoping property

The terms will cancel in a telescoping manner, which we can arrange into two columns for clarity:

\[\begin{align*} 2\sin\left(\frac{d}{2}\right)S = & \bcancel{ \sin\left(a + \frac{d}{2}\right)} - \sin\left(a - \frac{d}{2}\right) \\ & + \bcancel{\sin\left(a + \frac{3d}{2}\right)} - \bcancel{\sin\left(a + \frac{d}{2}\right)} \\ & + \bcancel{\sin\left(a + \frac{5d}{2}\right)} - \bcancel{\sin\left(a + \frac{3d}{2}\right)} \\ & + \ldots \\ & + \bcancel{\sin\left(a + (n-\frac{3}{2})d\right)} - \bcancel{\sin\left(a + (n-\frac{5}{2})d\right)} \\ & + \sin\left(a + (n-\frac{1}{2})d\right) - \bcancel{\sin\left(a + (n-\frac{3}{2})d\right)} \\ \end{align*}\]

\[ 2\sin\left(\frac{d}{2}\right)S = - \sin\left(a - \frac{d}{2}\right) + \sin\left(a + (n - \frac{1}{2})d\right) \]

Step 5: Simplify the remaining terms

Using the sum-to-product identity \( \sin C - \sin D = 2\sin\left(\frac{C-D}{2}\right)\cos\left(\frac{C+D}{2}\right) \):

\[ 2\sin\left(\frac{d}{2}\right)S = 2\sin\left(\frac{(a + (n - \frac{1}{2})d) - (a - \frac{d}{2})}{2}\right)\cos\left(\frac{(a + (n - \frac{1}{2})d) + (a - \frac{d}{2})}{2}\right) \]

\[ 2\sin\left(\frac{d}{2}\right)S = 2\sin\left(\frac{nd}{2}\right)\cos\left(a + \frac{(n - 1)d}{2}\right) \]

Step 6: Solve for \( S \)

Divide both sides by \( 2\sin\left(\frac{d}{2}\right) \) to isolate \( S \):

\[ S = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \cos\left(a + \frac{(n - 1)d}{2}\right) \]

And this completes the proof.

Alternate Proof

Although we have explored the sum of sine and cosine series through trigonometric identities, there exists an elegant alternative method involving complex numbers—a concept you will learn in more advanced mathematics. This method uses Euler's formula, which links complex exponentials to trigonometric functions, allowing the series of sines and cosines to be represented as a geometric series in the complex plane. In this geometric series, each term is a complex number whose magnitude and angle correspond to the sine and cosine terms we sum. The properties of geometric series make the calculation much more straightforward and lead to a compact formula for the sum. This approach offers a powerful example of how complex numbers can simplify and illuminate problems in trigonometry, providing a preview of the beautiful interconnectedness of different areas of mathematics that you will encounter in the future.

Alternating Series

Problem:

Find the sum of the series \( S = \sin(a) - \sin(a + d) + \sin(a + 2d) - \ldots + (-1)^{n-1}\sin(a + (n - 1)d) \).

Solution:

The sum of the alternating sine series can be computed by cleverly manipulating the series and then applying the formula for the sum of a sine series where the angles are in arithmetic progression.

Given the series:

\[ S = \sin(a) - \sin(a + d) + \sin(a + 2d) - \ldots + (-1)^{n-1}\sin(a + (n - 1)d) \]

Step 1: Rewrite the Series

We'll rewrite the series by incorporating \(\pi\) into the terms where the sine function is subtracted:

\[ S = \sin(a) + \sin(a + d + \pi) + \sin(a + 2d + 2\pi) + \ldots + \sin(a + (n - 1)d + (n - 1)\pi) \]

Because \(\sin(2n\pi + \theta) = \sin(\theta) \quad \text{and} \quad \sin((2n-1)\pi + \theta) = -\sin(\theta)\)

Step 2: Simplify the Series

\[ S = \sin(a) + \sin(a + (\pi + d)) + \sin(a + 2(\pi + d)) + \ldots + \sin(a + (n - 1)(\pi + d)) \]

Step 3: Apply the Standard Formula

Now, this is a standard sine series where the common difference is \(\pi + d\) instead of \(d\). The sum of a standard sine series where the angles are in arithmetic progression is given by:

\[ \sum_{k=0}^{n-1} \sin(a + kd) = \frac{\sin\left(\frac{nd}{2}\right)}{\sin\left(\frac{d}{2}\right)} \cdot \sin\left(a + \frac{(n - 1)d}{2}\right) \]

For our modified series, we replace \(d\) with \(\pi + d\) and apply the formula:

\[ S = \frac{\sin\left(\frac{n(\pi + d)}{2}\right)}{\sin\left(\frac{\pi + d}{2}\right)} \cdot \sin\left(a + \frac{(n - 1)(\pi + d)}{2}\right) \]

This is the sum of the series with alternating signs for the sine terms.

Product of COSINES of angles forming a GP with common ratio 2

For any angle \(\theta\),

\[ \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta} \]

Proof:

We begin the proof by multiplying and dividing the left-hand side (LHS) of the expression by \(2\sin \theta\):

\[ \text{LHS} = \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta \]

We multiply and divide by \(2\sin \theta\):

\[ \text{LHS} = \frac{2\sin \theta}{2\sin \theta} \cdot \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta \]

Now, we know from the double angle formula for sine that \(2\sin \theta \cos \theta = \sin 2\theta\). Applying this formula iteratively, we transform each term:

\[ \text{LHS} = \frac{\sin 2\theta}{2\sin \theta} \cdot \cos 2\theta \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta \]

\[ \text{LHS} = \frac{\sin 2\theta \cdot \cos 2\theta}{2\sin \theta} \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta \]

Applying the double angle formula again, we have \(\sin 2\theta \cdot \cos 2\theta = \frac{\sin 2^2\theta}{2}\):

\[ \text{LHS} = \frac{\sin 2^2\theta}{2^2 \sin \theta} \cdot \cos 2^2\theta \cdot \ldots \cdot \cos 2^{n-1}\theta \]

Continuing this process, we apply the double angle formula successively to each term in the product until we reach the last term:

\[ \text{LHS} = \frac{\sin 2^{n-1}\theta \cdot \cos 2^{n-1}\theta}{2^{n-1} \sin \theta} \]

For the final term, using the double angle formula one last time, we obtain \(\sin 2^{n-1}\theta \cdot \cos 2^{n-1}\theta = \frac{\sin 2^n\theta}{2}\):

\[ \text{LHS} = \frac{\sin 2^n\theta}{2^n \sin \theta} \]

This concludes the proof, showing that the product of cosines can indeed be expressed as the ratio of \(\sin 2^n \theta\) to \(2^n \sin \theta\).

Illustrations

The following illustrations demonstrate how to use the above formulas to find the values of certain series.

Example

Problem:

Prove that:

\[ \cos(\theta)\cos\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2^2}\right)\ldots\cos\left(\frac{\theta}{2^{n-1}}\right) = \frac{\sin (2\theta)}{2^n \sin \left(\frac{\theta}{2^{n-1}}\right)} \]

Solution:

We will utilize the proven formula:

\[ \cos \alpha \cdot \cos 2\alpha \cdot \cos 2^2\alpha \cdot \ldots \cdot \cos 2^{n-1}\alpha = \frac{\sin 2^n \alpha}{2^n \sin \alpha} \]

However, we need to manipulate our given product to match the form of the standard formula. We can reverse the series and replace \( \alpha \) in the formula with \( \frac{\theta}{2^{n-1}} \) so that the angles in the cosines match our series.

By reversing the series and replacing \( \alpha \) with \( \frac{\theta}{2^{n-1}} \), we have:

\[ \cos\left(\frac{\theta}{2^{n-1}}\right)\cos\left(\frac{\theta}{2^{n-2}}\right)\ldots\cos(\theta) \]

\[ = \cos \alpha \cdot \cos 2\alpha \cdot \cos 2^2\alpha \cdot \ldots \cdot \cos 2^{n-1}\alpha \]

Now, we can apply the standard formula and we get:

\[ \frac{\sin 2^n \left(\frac{\theta}{2^{n-1}}\right)}{2^n \sin \left(\frac{\theta}{2^{n-1}}\right)} \]

Simplifying, we get:

\[ \frac{\sin \left(\frac{2^n}{2^{n-1}}\theta\right)}{2^n \sin \left(\frac{\theta}{2^{n-1}}\right)} = \frac{\sin (2\theta)}{2^n \sin \left(\frac{\theta}{2^{n-1}}\right)} \]

Hence, the value of the product up to \( n \) terms is:

\[ \cos(\theta)\cos\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2^2}\right)\ldots\cos\left(\frac{\theta}{2^{n-1}}\right) = \frac{\sin (2\theta)}{2^n \sin \left(\frac{\theta}{2^{n-1}}\right)} \]

\(\blacksquare\)

Example

Problem:

Prove that:

\[ \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right) \]

is equal to \( \frac{1}{8} \).

Solution:

To find the value of \( \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right) \), we first convert \( \cos\left(\frac{3\pi}{7}\right) \) to \( \cos\left(\frac{4\pi}{7}\right) \) using the fact that \( \cos(\pi - \theta) = -\cos(\theta) \). So,

\[ \cos\left(\frac{3\pi}{7}\right) = -\cos\left(\pi - \frac{3\pi}{7}\right) = -\cos\left(\frac{4\pi}{7}\right) \]

Now we can express the product as:

\[ \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\left(-\cos\left(\frac{4\pi}{7}\right)\right) \]

This matches the form of the product in the proven formula with \( \theta = \frac{\pi}{7} \):

\[ \cos(\theta)\cos(2\theta)\cos(4\theta) = \frac{\sin(8\theta)}{8\sin(\theta)} \]

Substituting \( \theta = \frac{\pi}{7} \) into the formula gives us:

\[ \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\left(-\cos\left(\frac{4\pi}{7}\right)\right) = -\frac{\sin\left(8\cdot\frac{\pi}{7}\right)}{8\sin\left(\frac{\pi}{7}\right)} \]

Simplifying the sine term in the numerator,

\[ \sin\left(8\cdot\frac{\pi}{7}\right) = \sin\left(\pi + \frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right) \]

The \( \sin(\frac{\pi}{7}) \) in the numerator and denominator cancel out, so we're left with:

\[ \frac{\sin\left(\frac{\pi}{7}\right)}{8\sin\left(\frac{\pi}{7}\right)} = \frac{1}{8} \]

Therefore, the value of the product is:

\[ \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right) = \frac{1}{8} \]

In short,

\[ \begin{align*} \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{3\pi}{7}\right) &= \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\pi - \frac{4\pi}{7}\right) \\ &= \cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)(-\cos\left(\frac{4\pi}{7}\right)) \\ &= -\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right) \\ &= -\frac{\sin\left(8\cdot\frac{\pi}{7}\right)}{8\sin\left(\frac{\pi}{7}\right)} \\ &= -\frac{\sin\left(\pi + \frac{\pi}{7}\right)}{8\sin\left(\frac{\pi}{7}\right)} \\ &= \frac{\sin\left(\frac{\pi}{7}\right)}{8\sin\left(\frac{\pi}{7}\right)} \\ &= \frac{1}{8} \end{align*} \]

Example

Problem:

Prove that the product of cosines

\[ \cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\cos\left(\frac{5\pi}{11}\right) \]

is equal to \( \frac{1}{32} \).

Solution:

We group the terms and convert \( \cos\left(\frac{5\pi}{11}\right) \) to \( \cos\left(\frac{6\pi}{11}\right) \) using the identity \( \cos(\pi - \theta) = -\cos(\theta) \):

\[ \left[\cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\right]\left[\cos\left(\frac{3pi}{11}\right)\cos\left(\frac{5\pi}{11}\right)\right] \]

\[ = \left[\cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\right]\left[\cos\left(\frac{3pi}{11}\right)\cos\left(\pi - \frac{6\pi}{11}\right)\right] \]

\[ = \left[\cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\right]\left[\cos\left(\frac{3pi}{11}\right)\left(-\cos\left(\frac{6\pi}{11}\right)\right)\right] \]

Now, we apply the formula for the product of cosines:

For the first group, let \( \theta = \frac{\pi}{11} \) and \( n = 3 \):

\[ \cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right) = \frac{\sin\left(8\cdot\frac{\pi}{11}\right)}{8 \sin\left(\frac{\pi}{11}\right)} \]

For the second group, let \( \theta = \frac{3\pi}{11} \) and \( n = 2 \):

\[ \cos\left(\frac{3\pi}{11}\right)\cos\left(\frac{6\pi}{11}\right) = \frac{\sin\left(4\cdot\frac{3\pi}{11}\right)}{4 \sin\left(\frac{3\pi}{11}\right)} \]

Combining the two results:

\[ \frac{\sin\left(\frac{8\pi}{11}\right)}{8 \sin\left(\frac{\pi}{11}\right)} \cdot \left(-\frac{\sin\left(\frac{12\pi}{11}\right)}{4 \sin\left(\frac{3\pi}{11}\right)}\right) \]

Since \( \sin\left(\frac{8\pi}{11}\right) = \sin\left(\frac{3\pi}{11}\right) \) and \( \sin\left(\frac{12\pi}{11}\right) = -\sin\left(\frac{\pi}{11}\right) \) because \( \sin(\pi - x) = \sin(x) \) and \( \sin(\pi + x) = -\sin(x) \), we have:

\[ \frac{\sin\left(\frac{3\pi}{11}\right)}{8 \sin\left(\frac{\pi}{11}\right)} \cdot \left(-\frac{-\sin\left(\frac{\pi}{11}\right)}{4 \sin\left(\frac{3\pi}{11}\right)}\right) \]

The sines cancel out, leaving us with:

\[ \frac{1}{8} \cdot \frac{1}{4} = \frac{1}{32} \]

In short,

\[\begin{align*} \cos\left(\frac{\pi}{11}\right)&\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\cos\left(\frac{5\pi}{11}\right) \\ &= \cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)\cos\left(\pi - \frac{6\pi}{11}\right) \\ &= \cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\cos\left(\frac{3\pi}{11}\right)(-\cos\left(\frac{6\pi}{11}\right)) \\ &= \left[\cos\left(\frac{\pi}{11}\right)\cos\left(\frac{2\pi}{11}\right)\cos\left(\frac{4\pi}{11}\right)\right]\left[-\cos\left(\frac{3\pi}{11}\right)\cos\left(\frac{6\pi}{11}\right)\right] \\ &= \left[\frac{\sin\left(8\cdot\frac{\pi}{11}\right)}{8\sin\left(\frac{\pi}{11}\right)}\right]\left[-\frac{\sin\left(4\cdot\frac{3\pi}{11}\right)}{4\sin\left(\frac{3\pi}{11}\right)}\right] \\ &= \frac{\sin\left(\frac{8\pi}{11}\right)}{8\sin\left(\frac{\pi}{11}\right)} \cdot \frac{-\sin\left(\frac{12\pi}{11}\right)}{4\sin\left(\frac{3\pi}{11}\right)} \\ &= \frac{\sin\left(\pi-\frac{3\pi}{11}\right)}{8\sin\left(\frac{\pi}{11}\right)} \cdot \frac{-\sin\left(\pi + \frac{\pi}{11}\right)}{4\sin\left(\frac{3\pi}{11}\right)} \\ &= \frac{\sin\left(\frac{3\pi}{11}\right)}{8\sin\left(\frac{\pi}{11}\right)} \cdot \frac{\sin\left(\frac{\pi}{11}\right)}{4\sin\left(\frac{3\pi}{11}\right)} \\ &= \frac{1}{8} \cdot \frac{1}{4} \\ &= \frac{1}{32} \end{align*}\]