) 2 sin 15◦ cos 15◦ ; б) sin π/ 8 cos π/ 8 ; в) 4 sin 75◦ cos 75◦
To simplify these expressions, we can use the double angle identity for sine, which states that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta). \]
Now let's simplify each expression:
a) \(2 \sin 15^\circ \cos 15^\circ\)
Using the double angle identity:
\[ 2 \sin 15^\circ \cos 15^\circ = \sin(2 \times 15^\circ) = \sin 30^\circ. \]
Since \(\sin 30^\circ = \frac{1}{2}\), we have:
\[ 2 \sin 15^\circ \cos 15^\circ = \frac{1}{2}. \]
б) \(\sin \frac{\pi}{8} \cos \frac{\pi}{8}\)
Again, using the double angle identity:
\[ 2 \sin \frac{\pi}{8} \cos \frac{\pi}{8} = \sin\left(2 \cdot \frac{\pi}{8}\right) = \sin \frac{\pi}{4}. \]
Since \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\), we can write:
\[ \sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}. \]
в) \(4 \sin 75^\circ \cos 75^\circ\)
Using the double angle identity:
\[ 4 \sin 75^\circ \cos 75^\circ = 2 \cdot 2 \sin 75^\circ \cos 75^\circ = 2 \sin(2 \cdot 75^\circ) = 2 \sin 150^\circ. \]
Since \(\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}\), we have:
\[ 4 \sin 75^\circ \cos 75^\circ = 2 \cdot \frac{1}{2} = 1. \]
To summarize:
- a) \(2 \sin 15^\circ \cos 15^\circ = \frac{1}{2}\)
- б) \(\sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{\sqrt{2}}{4}\)
- в) \(4 \sin 75^\circ \cos 75^\circ = 1\)
