Question

Можно решение, если конечно не сложно. Ну или ссылку на rдз. Сам найти не могу почему-то.

Answer 1

1. $sin52^{\circ} = 2\cdot sin26^{\circ} \cdot cos26^{\circ}$

2. $cos\frac{4\pi}{5} = cos^2(\frac{2\pi}{5}) - sin^2(\frac{2\pi}{5})$

3. $tg64^{\circ} = \frac{2tg(32^{\circ})}{1 - tg^2(32^{\circ})}$

4. $sin(\frac{\pi }{2} - \alpha) = cos\alpha = cos^2(\frac\alpha2) - sin^2(\frac\alpha2)$

5. $cos(\pi - \alpha) = -cos\alpha = sin^2(\frac\alpha2) - cos^2(\frac\alpha2)$

6. $sin 6a = 2\cdot sin3\alpha \cdot cos3\alpha$

7. $cos7\alpha = cos^2(\frac{7\alpha}{2}) - sin^2(\frac{7\alpha}{2})$

8. $tg4\alpha = \frac{2tg(2\alpha)}{1-tg^2(2\alpha)}$

9. $sin\alpha \cdot cos\alpha = \frac{sin2\alpha}{2}$

10. $2sin\frac{5\pi }{12}cos\frac{5\pi}{12} = sin\frac{10\pi }{12} = sin\frac{5\pi}{6} = sin(\pi - \frac{\pi}{6}) = sin(\frac{\pi}{6}) = \frac{1}{2}$

11. $cos^2\frac{\pi}{12} - sin^2\frac{\pi}{12} = cos\frac{2\pi}{12} = cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$

12.

$(cos22,5^{\circ} - sin22,5^{\circ})^2 = cos^2(22,5^{\circ}) - 2sin(22,5^{\circ})cos(22,5^{\circ}) + sin^2(22,5^{\circ}) =\\\\ 1 - sin45^{\circ} = 1 - \frac{\sqrt2}{2} = 1 - 0,5\sqrt{2}$

Удачи!