Solving the Trigonometry Expression cos 80° + cos 40° - cos 20°

We need to find the value of the trigonometric expression \( \cos 80^\circ + \cos 40^\circ - \cos 20^\circ \). To simplify this expression, we can use trigonometric identities, specifically the sum-to-product formula.

Applying Sum-to-Product Identity

The sum-to-product identity for cosines is:

Let's apply this identity to the first two terms of the expression, \( \cos 80^\circ + \cos 40^\circ \). Here, \( A = 80^\circ \) and \( B = 40^\circ \).

Calculate the sum and difference of the angles:

• \( A+B = 80^\circ + 40^\circ = 120^\circ \)
• \( A-B = 80^\circ - 40^\circ = 40^\circ \)

Now, find the half-angles:

• \( \frac{A+B}{2} = \frac{120^\circ}{2} = 60^\circ \)
• \( \frac{A-B}{2} = \frac{40^\circ}{2} = 20^\circ \)

Substitute these values into the sum-to-product formula:

We know the exact value of \( \cos 60^\circ \). It is a standard angle value.

Using the table, \( \cos 60^\circ = \frac{1}{2} \). Substitute this into the expression:

Completing the Calculation

Now substitute this result back into the original expression:

Thus, the value of the expression \( \cos 80^\circ + \cos 40^\circ - \cos 20^\circ \) is 0.

Revision Table: Key Trigonometric Values

Remembering these standard values is crucial for solving many trigonometry problems.

Additional Information: Sum and Product Identities in Trigonometry

Sum-to-product and product-to-sum identities are useful for simplifying expressions or solving equations involving trigonometric functions. Here are a few related identities:

• Sum-to-Product:
• \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
• \( \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)
• \( \cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \) or \( 2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{B-A}{2} \right) \)
• Product-to-Sum:
• \( 2 \sin A \cos B = \sin(A+B) + \sin(A-B) \)
• \( 2 \cos A \sin B = \sin(A+B) - \sin(A-B) \)
• \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \)
• \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \)

These identities allow us to convert sums or differences of sines and cosines into products, and vice versa, which can simplify complex expressions like the one we solved.