Question

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

Answer 1

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

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

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

4. $sin(frac{pi }{2} - alpha) = cosalpha = cos^2(fracalpha2) - sin^2(fracalpha2)$

5. $cos(pi - alpha) = -cosalpha = sin^2(fracalpha2) - cos^2(fracalpha2)$

6. $sin 6a = 2cdot sin3alpha cdot cos3alpha$

7. $cos7alpha = cos^2(frac{7alpha}{2}) - sin^2(frac{7alpha}{2})$

8. $tg4alpha = frac{2tg(2alpha)}{1-tg^2(2alpha)}$

9. $sinalpha cdot cosalpha = frac{sin2alpha}{2}$

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

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

12.

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

Удачи!

Answer 2

Лучший)