Use \( \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \) to prove \( \sin (A+B) \) Prove \( \sin (A+B)-\sin (A-B)=2 \cos A \cdot \sin B \) Hence without a calculator prove that \( \sin 77^{\circ}-\sin 43^{\circ}=\operatorname{Sin} 17^{\circ} \)

To prove \( \sin (A + B) - \sin (A - B) = 2 \cos A \cdot \sin B \), we start by expressing \( \sin (A + B) \) and \( \sin (A - B) \) using the sine addition and subtraction formulas. We know: \[ \sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B \] \[ \sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B \] Now, subtract the second equation from the first: \[ \sin (A + B) - \sin (A - B) = (\sin A \cdot \cos B + \cos A \cdot \sin B) - (\sin A \cdot \cos B - \cos A \cdot \sin B) \] This simplifies to: \[ = 2 \cos A \cdot \sin B \] Now, using this proven identity, we can substitute specific angles. Let \( A = 60^{\circ} \) and \( B = 17^{\circ} \): Using our identity gives: \[ \sin (60^{\circ} + 17^{\circ}) - \sin (60^{\circ} - 17^{\circ}) = 2 \cos 60^{\circ} \cdot \sin 17^{\circ} \] Calculating the left side: \[ \sin 77^{\circ} - \sin 43^{\circ} \] Calculating the right side: \[ 2 \cdot \frac{1}{2} \cdot \sin 17^{\circ} = \sin 17^{\circ} \] This means: \[ \sin 77^{\circ} - \sin 43^{\circ} = \sin 17^{\circ} \] Thus we have proven the statement! Lastly, a fun fact to keep in mind is that the sine function is periodic, meaning the values repeat every \(360^\circ\). So when you’re dealing with angles, it can help to think about simplifying down to acute angles or recognizable values when you’re adding or subtracting angles. Now, speaking of applications, trigonometry isn’t just about finding angles in math classes; it’s widely used in construction, engineering, and even in computer graphics! Understanding how to manipulate and work with these sine and cosine functions can aid in crafting precise structures or even creating lifelike animations in video games.