7.1. Calculate: $4\left(\sin ^{3} \frac{49 \pi}{48} \cos \frac{49 \pi}{16}+\cos ^{3} \frac{49 \pi}{48} \sin \frac{49 \pi}{16}\right) \cos \frac{49 \pi}{12}$.
Answer: 0.75. Solution: Let $\alpha=\frac{49 \pi}{48}=\pi+\frac{\pi}{48}$. Then:

$\left(\sin ^{3} \alpha \cos 3 \alpha+\cos ^{3} \alpha \sin 3 \alpha\right) \cos 4 \alpha=\left[\sin ^{3} \alpha\left(4 \cos ^{3} \alpha-3 \cos \alpha\right)+\cos ^{3} \alpha\left(3 \sin \alpha-4 \sin ^{3} \alpha\right)\right] \cos 4 \alpha$ $=\left(-3 \sin ^{3} \alpha \cos \alpha+3 \cos ^{3} \alpha \sin \alpha\right) \cos 4 \alpha=3 \sin \alpha \cos \alpha\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right) \cos 4 \alpha=\frac{3}{2} \sin 2 \alpha \cos 2 \alpha \cos 4 \alpha$ $=\frac{3}{4} \sin 4 \alpha \cos 4 \alpha=\frac{3}{8} \sin 8 \alpha$. Therefore, the desired number is: $4 \cdot \frac{3}{8} \sin \left(8 \pi+\frac{8 \pi}{48}\right)=\frac{3}{2} \cdot \sin \frac{\pi}{6}=\frac{3}{4}=0.75$.