Evaluate the trigonometric expression 2 cos(π/13) cos(9π/13) + cos(3π/13) + cos(5π/13).

Trigonometric Expression Evaluation

Evaluate the expression:

\[ 2 \cos\left(\frac{\pi}{13}\right) \cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \]

Select the correct option:

a) -1

b) 0

c) 1

d) none of the above

Detailed Solution

Step 1: Write Down the Expression

We need to evaluate:

\[ 2 \cos\left(\frac{\pi}{13}\right) \cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \]

Step 2: Identify Useful Identities

Notice that the first term is a product of two cosines multiplied by 2. This reminds us of a trigonometric identity that converts a product into a sum. The identity we’ll use is:

\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]

This is super handy because it transforms the product into a sum of cosines, matching the form of the other terms in the expression.

Step 3: Apply the Identity

Let’s apply this to the first term. Set:

\[ A = \frac{\pi}{13}, \quad B = \frac{9\pi}{13} \]

Now compute the arguments:

\[ A + B = \frac{\pi}{13} + \frac{9\pi}{13} = \frac{10\pi}{13} \] \[ A - B = \frac{\pi}{13} - \frac{9\pi}{13} = \frac{\pi - 9\pi}{13} = \frac{-8\pi}{13} \]

Since cosine is an even function (\(\cos(-\theta) = \cos(\theta)\)), we have:

\[ \cos\left(\frac{-8\pi}{13}\right) = \cos\left(\frac{8\pi}{13}\right) \]

So:

\[ 2 \cos\left(\frac{\pi}{13}\right) \cos\left(\frac{9\pi}{13}\right) = \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right) \]

Step 4: Substitute Back into the Expression

Replace the first term in the original expression:

\[ \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \]

Now we have a sum of four cosine terms. Our goal is to simplify this sum and find its value.

Step 5: Look for Patterns or Relationships

Let’s list the angles: \(\frac{3\pi}{13}\), \(\frac{5\pi}{13}\), \(\frac{8\pi}{13}\), \(\frac{10\pi}{13}\). Do they relate to each other in a way that simplifies the sum? A useful property of cosine is that \(\cos(\pi - \theta) = -\cos(\theta)\). Let’s check if any pairs add up to \(\pi\):

• \(\frac{3\pi}{13} + \frac{10\pi}{13} = \frac{13\pi}{13} = \pi\)
• \(\frac{5\pi}{13} + \frac{8\pi}{13} = \frac{13\pi}{13} = \pi\)

Wow, that’s promising! If two angles sum to \(\pi\), their cosines are negatives of each other:

\[ \cos\left(\frac{10\pi}{13}\right) = \cos\left(\pi - \frac{3\pi}{13}\right) = -\cos\left(\frac{3\pi}{13}\right) \] \[ \cos\left(\frac{8\pi}{13}\right) = \cos\left(\pi - \frac{5\pi}{13}\right) = -\cos\left(\frac{5\pi}{13}\right) \]

Step 6: Simplify the Sum

Substitute these into the expression:

\[ \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right)\]= \[-\cos\left(\frac{3\pi}{13}\right) + -\cos\left(\frac{5\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \]

Group the terms:

\[ [-\cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right)] + [-\cos\left(\frac{5\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right)] = 0 + 0 = 0 \]

The terms cancel out perfectly! So:

\[ \cos\left(\frac{10\pi}{13}\right) + \cos\left(\frac{8\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) = 0 \]

Since this equals the original expression, we have:

\[ 2 \cos\left(\frac{\pi}{13}\right) \cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) = 0 \]

Final Answer

The value of the expression is 0, so the correct option is b) 0.

Great job following along! Practice similar problems to master these identities—they’re key for your exams!