Evaluating the Trigonometric Expression cos 20° cos 40° cos 80°

The problem asks us to evaluate the product of three cosine terms: $\cos 20^\circ$, $\cos 40^\circ$, and $\cos 80^\circ$. This type of expression, where the angles are in a geometric progression with a common ratio of 2, often involves using the double angle formula for sine, $\sin 2A = 2 \sin A \cos A$, or its rearrangement, $\cos A = \frac{\sin 2A}{2 \sin A}$.

Step-by-Step Solution for cos 20° cos 40° cos 80°

Let the given expression be $E$.

$$E = \cos 20^\circ \cos 40^\circ \cos 80^\circ$$

We can introduce a sine term to use the double angle identity. Let's multiply and divide the expression by $2 \sin 20^\circ$:

$$E = \frac{1}{2 \sin 20^\circ} (2 \sin 20^\circ \cos 20^\circ) \cos 40^\circ \cos 80^\circ$$

Using the identity $2 \sin A \cos A = \sin 2A$ with $A = 20^\circ$, we get:

$$2 \sin 20^\circ \cos 20^\circ = \sin (2 \times 20^\circ) = \sin 40^\circ$$

Substitute this back into the expression for $E$:

$$E = \frac{1}{2 \sin 20^\circ} (\sin 40^\circ) \cos 40^\circ \cos 80^\circ$$

Now, we have $\sin 40^\circ \cos 40^\circ$. We can use the double angle identity again. Multiply the term $(\sin 40^\circ \cos 40^\circ)$ by 2 and divide by 2:

$$E = \frac{1}{2 \sin 20^\circ} \left( \frac{1}{2} (2 \sin 40^\circ \cos 40^\circ) \right) \cos 80^\circ$$

Using the identity $2 \sin A \cos A = \sin 2A$ with $A = 40^\circ$, we get:

$$2 \sin 40^\circ \cos 40^\circ = \sin (2 \times 40^\circ) = \sin 80^\circ$$

Substitute this back into the expression for $E$:

$$E = \frac{1}{2 \sin 20^\circ} \left( \frac{1}{2} \sin 80^\circ \right) \cos 80^\circ$$

$$E = \frac{1}{4 \sin 20^\circ} \sin 80^\circ \cos 80^\circ$$

Again, we have $\sin 80^\circ \cos 80^\circ$. Use the double angle identity one more time. Multiply the term $(\sin 80^\circ \cos 80^\circ)$ by 2 and divide by 2:

$$E = \frac{1}{4 \sin 20^\circ} \left( \frac{1}{2} (2 \sin 80^\circ \cos 80^\circ) \right)$$

Using the identity $2 \sin A \cos A = \sin 2A$ with $A = 80^\circ$, we get:

$$2 \sin 80^\circ \cos 80^\circ = \sin (2 \times 80^\circ) = \sin 160^\circ$$

Substitute this back into the expression for $E$:

$$E = \frac{1}{4 \sin 20^\circ} \left( \frac{1}{2} \sin 160^\circ \right)$$

$$E = \frac{\sin 160^\circ}{8 \sin 20^\circ}$$

Now, we need to simplify $\sin 160^\circ$. We know that $\sin (180^\circ - \theta) = \sin \theta$. Using this property:

$$\sin 160^\circ = \sin (180^\circ - 20^\circ) = \sin 20^\circ$$

Substitute this into the expression for $E$:

$$E = \frac{\sin 20^\circ}{8 \sin 20^\circ}$$

Since $\sin 20^\circ \neq 0$, we can cancel the $\sin 20^\circ$ terms:

$$E = \frac{1}{8}$$

Thus, the value of $\cos 20^\circ \cos 40^\circ \cos 80^\circ$ is $\frac{1}{8}$.

Understanding the Product of Cosines

The technique used here is a general method for evaluating products of cosines where the angles are in geometric progression ($\theta, 2\theta, 4\theta, \dots, 2^{n-1}\theta$). The general formula is:

$$\cos \theta \cos 2\theta \cos 4\theta \dots \cos (2^{n-1}\theta) = \frac{\sin (2^n \theta)}{2^n \sin \theta}$$

In this problem, $\theta = 20^\circ$ and $n = 3$ (since $2^{3-1} \times 20^\circ = 4 \times 20^\circ = 80^\circ$, which is the last term). Applying the formula directly:

$$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin (2^3 \times 20^\circ)}{2^3 \sin 20^\circ} = \frac{\sin (8 \times 20^\circ)}{8 \sin 20^\circ} = \frac{\sin 160^\circ}{8 \sin 20^\circ}$$

Since $\sin 160^\circ = \sin (180^\circ - 20^\circ) = \sin 20^\circ$, we get:

$$\frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$$

Both methods yield the same result, $\frac{1}{8}$.

Revision Table: Key Concepts in Evaluating cos 20 cos 40 cos 80

• The problem involves evaluating a product of cosines with angles in a specific pattern (geometric progression).
• The key identities used are the double angle formula for sine ($\sin 2A = 2 \sin A \cos A$) and the property of sine of supplementary angles ($\sin (180^\circ - A) = \sin A$).
• By repeatedly applying the double angle formula after introducing a $\sin \theta$ term, the product can be simplified.
• There is a general formula for the product of cosines $\cos \theta \cos 2\theta \dots \cos (2^{n-1}\theta)$ which can also be used.

Additional Information on Trigonometric Product Formulas

Evaluating products of trigonometric functions like $\cos \theta \cos 2\theta \cos 4\theta$ is a common problem in trigonometry. The method demonstrated is versatile and can be applied to similar products. For instance, $\cos 10^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ$ can be evaluated using the same approach or the general formula. The general formula is derived by repeatedly using the identity $\cos A = \frac{\sin 2A}{2 \sin A}$:

$\cos \theta = \frac{\sin 2\theta}{2 \sin \theta}$

$\cos \theta \cos 2\theta = \frac{\sin 2\theta}{2 \sin \theta} \cos 2\theta = \frac{1}{2 \sin \theta} (\sin 2\theta \cos 2\theta) = \frac{1}{2 \sin \theta} \left( \frac{\sin 4\theta}{2} \right) = \frac{\sin 4\theta}{4 \sin \theta}$

$\cos \theta \cos 2\theta \cos 4\theta = \frac{\sin 4\theta}{4 \sin \theta} \cos 4\theta = \frac{1}{4 \sin \theta} (\sin 4\theta \cos 4\theta) = \frac{1}{4 \sin \theta} \left( \frac{\sin 8\theta}{2} \right) = \frac{\sin 8\theta}{8 \sin \theta}$

Continuing this pattern $n$ times leads to the general formula $\frac{\sin (2^n \theta)}{2^n \sin \theta}$. This highlights how the step-by-step method is essentially a derivation of this general formula in a specific case.