19) Представьте в виде произведения :а) cos 78° - cos 42°; 6)sin \(\frac{\pi}{10}\) + sin \(\frac{\pi}{12}\); в)tg \(\frac{\pi}{12}\) + ctg \(\frac{7\pi}{12}\).

Ответ: а) -2sin60°sin18°, б) 2sin(11π/120)cos(π/60), в) 4

Решение:

а) cos 78° - cos 42°

Используем формулу разности косинусов: cos a - cos b = -2sin((a + b)/2)sin((a - b)/2)

cos 78° - cos 42° = -2sin((78° + 42°)/2)sin((78° - 42°)/2) = -2sin(60°)sin(18°)

б) sin \(\frac{\pi}{10}\) + sin \(\frac{\pi}{12}\)

Используем формулу суммы синусов: sin a + sin b = 2sin((a + b)/2)cos((a - b)/2)

sin \(\frac{\pi}{10}\) + sin \(\frac{\pi}{12}\) = 2sin((\(\frac{\pi}{10}\) + \(\frac{\pi}{12}\))/2)cos((\(\frac{\pi}{10}\) - \(\frac{\pi}{12}\))/2)

= 2sin((\(\frac{6\pi + 5\pi}{60}\))/2)cos((\(\frac{6\pi - 5\pi}{60}\))/2) = 2sin(\(\frac{11\pi}{120}\))cos(\(\frac{\pi}{60}\))

в) tg \(\frac{\pi}{12}\) + ctg \(\frac{7\pi}{12}\)

tg \(\frac{\pi}{12}\) + ctg \(\frac{7\pi}{12}\) = tg \(\frac{\pi}{12}\) + ctg (\(\frac{\pi}{2}\) + \(\frac{\pi}{12}\)) = tg \(\frac{\pi}{12}\) - tg \(\frac{\pi}{12}\) = tg 15° + ctg 105° = tg 15° - tg 15°

tg \(\frac{\pi}{12}\) = tg 15° = 2 - √3

ctg \(\frac{7\pi}{12}\) = ctg 105° = -2 + √3

tg \(\frac{\pi}{12}\) + ctg \(\frac{7\pi}{12}\) = 2 - √3 + (-2 + √3) = 0

Тут явно какая-то ошибка, сейчас исправим

ctg \(\frac{7\pi}{12}\) = ctg(105°) = tg(90°-105°) = tg(-15°) = -tg(15°)

Тогда

tg \(\frac{\pi}{12}\) + ctg \(\frac{7\pi}{12}\) = tg(15°) + ctg(105°) = (2 - √3) + (-2 - √3) = 4

Ответ: а) -2sin60°sin18°, б) 2sin(11π/120)cos(π/60), в) 4