Найти значение острого угла α, если: 1) cos 75° = cos (90° – α); 3) sin 150° = sin (180° – α); 5) sin 5/4 π = sin (π + α); 7) cos 7/4 π = cos (3/2 π + α); 2) sin 150° = sin (90° + α); 4) cos 310° = cos (270° + α); 6) tg π/5 = tg (π/2 - α); 8) ctg 11/6 π = ctg (2π-α). Вычислить с помощью формулы приведения (525-526). 525 1) cos 150°; 2) sin 135°; 3) ctg 135°; 4) cos 120°; 5) cos 225°; 6) sin 210°; 7) ctg 240°; 8) sin 315°. 526 1) tg 5π/4; 2) sin 7π/6; 3) cos 5π/3; 4) ctg 5π/3; 5) sin (-13π/6); 6) cos (-7π/3); 7) tg (-2π/3); 8) ctg (-7π/4)

Ответ: Решения ниже

Задание 524

1) cos 75° = cos (90° – α)

90 - α = 75

α = 90 - 75

α = 15°

2) sin 150° = sin (90° + α)

90 + α = 150

α = 150 - 90

α = 60°

3) sin 150° = sin (180° – α)

180 - α = 150

α = 180 - 150

α = 30°

4) cos 310° = cos (270° + α)

270 + α = 310

α = 310 - 270

α = 40°

5) sin \(\frac{5}{4}\) π = sin (π + α)

π + α = \(\frac{5}{4}\) π

α = \(\frac{5}{4}\) π - π

α = \(\frac{5}{4}\) π - \(\frac{4}{4}\) π

α = \(\frac{1}{4}\) π

6) tg \(\frac{π}{5}\) = tg (\(\frac{π}{2}\) - α)

\(\frac{π}{2}\) - α = \(\frac{π}{5}\)

α = \(\frac{π}{2}\) - \(\frac{π}{5}\)

α = \(\frac{5π}{10}\) - \(\frac{2π}{10}\)

α = \(\frac{3π}{10}\)

7) cos \(\frac{7π}{4}\) = cos (\(\frac{3}{2}\) π + α)

\(\frac{3}{2}\) π + α = \(\frac{7π}{4}\)

α = \(\frac{7π}{4}\) - \(\frac{3}{2}\) π

α = \(\frac{7π}{4}\) - \(\frac{6π}{4}\)

α = \(\frac{π}{4}\)

8) ctg \(\frac{11}{6}\) π = ctg (2π-α)

2π - α = \(\frac{11}{6}\) π

α = 2π - \(\frac{11}{6}\) π

α = \(\frac{12π}{6}\) - \(\frac{11}{6}\) π

α = \(\frac{π}{6}\)

Задание 525

1) cos 150° = -cos (180 - 150) = -cos 30 = -\(\frac{\sqrt{3}}{2}\)

2) sin 135° = sin (180 - 45) = sin 45 = \(\frac{\sqrt{2}}{2}\)

3) ctg 135° = -ctg (180 - 45) = -ctg 45 = -1

4) cos 120° = -cos (180 - 60) = -cos 60 = -\(\frac{1}{2}\)

5) cos 225° = -cos (225 - 180) = -cos 45 = -\(\frac{\sqrt{2}}{2}\)

6) sin 210° = -sin (210 - 180) = -sin 30 = -\(\frac{1}{2}\)

7) ctg 240° = ctg (240 - 180) = ctg 60 = \(\frac{\sqrt{3}}{3}\)

8) sin 315° = -sin (360 - 315) = -sin 45 = -\(\frac{\sqrt{2}}{2}\)

Задание 526

1) tg \(\frac{5π}{4}\) = tg (\(π + \frac{π}{4}\)) = tg \(\frac{π}{4}\) = 1

2) sin \(\frac{7π}{6}\) = -sin (\(\frac{7π}{6}\) - π) = -sin \(\frac{π}{6}\) = -\(\frac{1}{2}\)

3) cos \(\frac{5π}{3}\) = cos (2π - \(\frac{π}{3}\)) = cos \(\frac{π}{3}\) = \(\frac{1}{2}\)

4) ctg \(\frac{5π}{3}\) = -ctg (2π - \(\frac{π}{3}\)) = -ctg \(\frac{π}{3}\) = -\(\frac{\sqrt{3}}{3}\)

5) sin (\(-\frac{13π}{6}\)) = -sin (\(\frac{13π}{6}\)) = -sin (2π + \(\frac{π}{6}\)) = -sin \(\frac{π}{6}\) = -\(\frac{1}{2}\)

6) cos (\(-\frac{7π}{3}\)) = cos (\(\frac{7π}{3}\)) = cos (2π + \(\frac{π}{3}\)) = cos \(\frac{π}{3}\) = \(\frac{1}{2}\)

7) tg (\(-\frac{2π}{3}\)) = -tg (\(\frac{2π}{3}\)) = tg (π - \(\frac{π}{3}\)) = -tg \(\frac{π}{3}\) = -\(\sqrt{3}\)

8) ctg (\(-\frac{7π}{4}\)) = -ctg (\(\frac{7π}{4}\)) = -ctg (2π - \(\frac{π}{4}\)) = ctg \(\frac{π}{4}\) = 1

Ответ: Решения выше