Evaluating a Trigonometric Product

We are asked to find the value of the trigonometric expression:

\(\left( {1 + \cos \frac{{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{3{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{5{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{7{\rm{\pi }}}{8}} \right)\)

Understanding the Angles

Let's examine the angles in the expression: \(\frac{{\rm{\pi }}}{8}\), \(\frac{{3{\rm{\pi }}}}{8}\), \(\frac{{5{\rm{\pi }}}}{8}\), and \(\frac{{7{\rm{\pi }}}}{8}\). We can notice relationships between them:

• \(\frac{{5{\rm{\pi }}}}{8} = \pi - \frac{{3{\rm{\pi }}}}{8}\)
• \(\frac{{7{\rm{\pi }}}}{8} = \pi - \frac{{\rm{\pi }}}{8}\)

Using Trigonometric Identities

We can use the identity \(\cos(\pi - x) = -\cos x\). Applying this to the third and fourth terms:

• \(\cos \frac{{5{\rm{\pi }}}}{8} = \cos \left(\pi - \frac{{3{\rm{\pi }}}}{8}\right) = -\cos \frac{{3{\rm{\pi }}}}{8}\)
• \(\cos \frac{{7{\rm{\pi }}}}{8} = \cos \left(\pi - \frac{{\rm{\pi }}}{8}\right) = -\cos \frac{{\rm{\pi }}}{8}\)

Now substitute these back into the original expression. Let the expression be \(P\).

\(P = \left( {1 + \cos \frac{{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{3{\rm{\pi }}}}{8}} \right)\left( {1 + \left(-\cos \frac{{3{\rm{\pi }}}}{8}\right)} \right)\left( {1 + \left(-\cos \frac{{\rm{\pi }}}{8}\right)} \right)\)

\(P = \left( {1 + \cos \frac{{\rm{\pi }}}{8}} \right)\left( {1 + \cos \frac{{3{\rm{\pi }}}}{8}} \right)\left( {1 - \cos \frac{{3{\rm{\pi }}}}{8}} \right)\left( {1 - \cos \frac{{\rm{\pi }}}{8}} \right)\)

Rearranging and Applying Difference of Squares

Let's rearrange the terms to group them conveniently:

\(P = \left[ \left( {1 + \cos \frac{{\rm{\pi }}}{8}} \right)\left( {1 - \cos \frac{{\rm{\pi }}}{8}} \right) \right] \left[ \left( {1 + \cos \frac{{3{\rm{\pi }}}}{8}} \right)\left( {1 - \cos \frac{{3{\rm{\pi }}}}{8}} \right) \right]\)

Now, we use the difference of squares identity: \((a+b)(a-b) = a^2 - b^2\)

\(P = \left( 1^2 - \cos^2 \frac{{\rm{\pi }}}{8} \right) \left( 1^2 - \cos^2 \frac{{3{\rm{\pi }}}}{8} \right)\)

\(P = \left( 1 - \cos^2 \frac{{\rm{\pi }}}{8} \right) \left( 1 - \cos^2 \frac{{3{\rm{\pi }}}}{8} \right)\)

Using Pythagorean Identity

Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\), which implies \(1 - \cos^2 x = \sin^2 x\).

Applying this identity:

\(P = \left( \sin^2 \frac{{\rm{\pi }}}{8} \right) \left( \sin^2 \frac{{3{\rm{\pi }}}}{8} \right)\)

\(P = \sin^2 \frac{{\rm{\pi }}}{8} \sin^2 \frac{{3{\rm{\pi }}}}{8}\)

Using Complementary Angle Identity

Notice that \(\frac{{3{\rm{\pi }}}}{8} = \frac{{4{\rm{\pi }}}}{8} - \frac{{\rm{\pi }}}{8} = \frac{{\rm{\pi }}}{2} - \frac{{\rm{\pi }}}{8}\).

We can use the complementary angle identity: \(\sin(\frac{{\rm{\pi }}}{2} - x) = \cos x\).

\(\sin \frac{{3{\rm{\pi }}}}{8} = \sin \left(\frac{{\rm{\pi }}}{2} - \frac{{\rm{\pi }}}{8}\right) = \cos \frac{{\rm{\pi }}}{8}\)

Substitute this back into the expression for \(P\):

\(P = \sin^2 \frac{{\rm{\pi }}}{8} \left( \cos \frac{{\rm{\pi }}}{8} \right)^2\)

\(P = \sin^2 \frac{{\rm{\pi }}}{8} \cos^2 \frac{{\rm{\pi }}}{8}\)

\(P = \left( \sin \frac{{\rm{\pi }}}{8} \cos \frac{{\rm{\pi }}}{8} \right)^2\)

Using Double Angle Identity

Recall the double angle identity for sine: \(\sin 2x = 2 \sin x \cos x\), which means \(\sin x \cos x = \frac{1}{2} \sin 2x\).

Applying this identity:

\(P = \left( \frac{1}{2} \sin \left( 2 \cdot \frac{{\rm{\pi }}}{8} \right) \right)^2\)

\(P = \left( \frac{1}{2} \sin \frac{{2{\rm{\pi }}}}{8} \right)^2\)

\(P = \left( \frac{1}{2} \sin \frac{{\rm{\pi }}}{4} \right)^2\)

Evaluating the Final Term

The value of \(\sin \frac{{\rm{\pi }}}{4}\) is \(\frac{1}{\sqrt{2}}\).

\(P = \left( \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \right)^2\)

\(P = \left( \frac{1}{2\sqrt{2}} \right)^2\)

Now, square the term:

\(P = \frac{1^2}{(2\sqrt{2})^2} = \frac{1}{2^2 \cdot (\sqrt{2})^2} = \frac{1}{4 \cdot 2} = \frac{1}{8}\)

Thus, the value of the given expression is \(\frac{1}{8}\).

Summary of Steps

Revision Table: Key Trigonometric Identities

Additional Information on Trigonometric Product Evaluation

Evaluating products involving trigonometric functions of angles in arithmetic progression often involves using specific identities to simplify the expression. In this case, the angles \(\frac{{\rm{\pi }}}{8}\), \(\frac{{3{\rm{\pi }}}}{8}\), \(\frac{{5{\rm{\pi }}}}{8}\), \(\frac{{7{\rm{\pi }}}}{8}\) are in an arithmetic progression with a common difference of \(\frac{{2{\rm{\pi }}}}{8} = \frac{{\rm{\pi }}}{4}\). However, the key to simplifying this specific product was recognizing the supplementary angle relationship (\(\frac{{5\pi}}{8} = \pi - \frac{{3\pi}}{8}\), \(\frac{{7\pi}}{8} = \pi - \frac{{\rm{\pi }}}{8}\)) which allowed pairing terms using the difference of squares identity. This reduced the problem to a product of squares of sine terms. The complementary angle relationship (\(\frac{{3\pi}}{8} = \frac{{\rm{\pi }}}{2} - \frac{{\rm{\pi }}}{8}\)) then transformed one sine term into a cosine term, leading to the form \(\sin^2 x \cos^2 x\). This form is directly related to the sine double angle formula, enabling final evaluation. Problems like this highlight the importance of being familiar with various trigonometric identities and recognizing angle relationships.