Problem 91 - Algebra and Trigonometry

Question

Show that \(\sin 130^{\circ}-\sin 110^{\circ}=-\sin 10^{\circ}\).

Step-by-Step Solution

Answer

The identity holds because \(\sin 130^{\circ} - \sin 110^{\circ} = -\sin 10^{\circ}\).

Step 1: Apply Sine Difference Formula

We start with the formula for the difference of two sines: \[\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)\]Substituting \(A = 130^{\circ}\) and \(B = 110^{\circ}\) into the formula, we get:\[\sin 130^{\circ} - \sin 110^{\circ} = 2 \cos\left(\frac{130^{\circ} + 110^{\circ}}{2}\right) \sin\left(\frac{130^{\circ} - 110^{\circ}}{2}\right)\]

Step 2: Simplify the Angles in the Formula

Calculate the angles in the expressions:\[\frac{130^{\circ} + 110^{\circ}}{2} = \frac{240^{\circ}}{2} = 120^{\circ}\]\[\frac{130^{\circ} - 110^{\circ}}{2} = \frac{20^{\circ}}{2} = 10^{\circ}\]

Step 3: Substitute Back into the Formula

Substitute these simplifications back into the formula:\[\sin 130^{\circ} - \sin 110^{\circ} = 2 \cos(120^{\circ}) \sin(10^{\circ})\]

Step 4: Evaluate the Expression

We know from trigonometric values that:\[\cos(120^{\circ}) = -\frac{1}{2}\]Substitute this into the expression:\[2 \cos(120^{\circ}) \sin(10^{\circ}) = 2 \left(-\frac{1}{2}\right) \sin(10^{\circ}) = -\sin(10^{\circ})\]

Step 5: Conclusion

Thus, we have shown that:\[\sin 130^{\circ} - \sin 110^{\circ} = -\sin 10^{\circ}\]

Key Concepts

Understanding the Sine Difference Formula

The sine difference formula is a powerful tool in trigonometry that simplifies the difference between the sines of two angles. It is especially useful when dealing with complex angle problems. The formula is expressed as:

• \( \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \)

This equation helps us transform a subtraction problem into a multiplication, making it easier to solve. By converting the sine difference into a product of sine and cosine, we can leverage known trigonometric values to find our answer. In the given exercise, substituting \(A = 130^{\circ}\) and \(B = 110^{\circ}\) allows us to proceed with simplifying without directly calculating large angle sines. This method often reveals elegant solutions hidden in otherwise complex-looking trigonometric expressions.

Angle Simplification in Trigonometric Formulas

To apply the sine difference formula effectively, the angles within the formula need to be simplified. This involves two parts:

• The average of the angles \( \frac{A + B}{2} \)
• The difference divided by two \( \frac{A - B}{2} \)

For example, in our case:

• The average of \(130^{\circ}\) and \(110^{\circ}\) is \(\frac{240^{\circ}}{2} = 120^{\circ}\)
• The difference is \(\frac{20^{\circ}}{2} = 10^{\circ}\)

These simplifications turn an otherwise tedious process of calculating the sine values directly into a manageable task using known trigonometric values. The angles thus simplified assist in accurate calculation while also making demonstration of identities like the one in this exercise clear and straightforward.

Trigonometric Values and Their Usefulness

Trigonometric values are the backbone of solving trigonometric equations and identities efficiently. These include well-known values for sine and cosine at particular angles, such as 0°, 30°, 45°, 60°, 90°, and so on. In our exercise, utilizing the trigonometric value \( \cos(120^{\circ}) = -\frac{1}{2} \) is crucial. This knowledge allows us to substitute directly into our sine difference formula:- \( 2 \cos(120^{\circ}) \sin(10^{\circ}) = 2 \left(-\frac{1}{2}\right) \sin(10^{\circ}) = -\sin(10^{\circ}) \)This simple substitution turns the problem into a straightforward calculation, enhancing both comprehension and speed. Familiarity with these trigonometric values means students can tackle a broad range of problems they might encounter in their studies, making them less reliant on calculators for solutions and more proficient in recognizing patterns in trigonometric identities.