Самостоятельная работа: sin2d, sin I bapieasem Вычисленте: a) 2 sin 15°c03150 51. (cos.75°-sin 750)2 21. sin.cos+/7/ 6). 2 sin cos • Известно, что Sint=万,麥LLLT. Haigume: a). Sin 2t 5). cosat 6) Eg 2t e) ctg 2 t Вадано: cost=104L 4 cos; sint忘;ty责;olg Вычи асесть COS б) Дено: ctgt=45냐 ㄴ 올 Вычислить: Cos责; sin贵;ty责;ctg ④. Упростите выражение. a) sint d cos 2 &t cost 8). CDs+in COS

Ответ: Решения ниже

1. Вычислите:

a) 2 sin 15° cos 15°

Используем формулу синуса двойного угла: sin 2α = 2 sin α cos α

Тогда: 2 sin 15° cos 15° = sin (2 ⋅ 15°) = sin 30° = 1/2

б) 2 sin (π/7) cos (π/7)

Аналогично: 2 sin (π/7) cos (π/7) = sin (2π/7)

в) sin (π/8) cos (π/8) + 1/4

sin (π/8) cos (π/8) = 1/2 ⋅ 2 sin (π/8) cos (π/8) = 1/2 sin (2π/8) = 1/2 sin (π/4) = 1/2 ⋅ \(\frac{\sqrt{2}}{2}\) = \(\frac{\sqrt{2}}{4}\)

Тогда: sin (π/8) cos (π/8) + 1/4 = \(\frac{\sqrt{2}}{4}\) + 1/4 = \(\frac{\sqrt{2} + 1}{4}\)

г) (cos 75° - sin 75°)²

(cos 75° - sin 75°)² = cos² 75° - 2 cos 75° sin 75° + sin² 75° = 1 - sin (2 ⋅ 75°) = 1 - sin 150° = 1 - 1/2 = 1/2

2. Известно, что sin t = 5/13, π/2 < t < π. Найдите:

a) sin 2t

cos t = -\(\sqrt{1 - sin^2 t}\) = -\(\sqrt{1 - (5/13)^2}\) = -\(\sqrt{1 - 25/169}\) = -\(\sqrt{144/169}\) = -12/13

sin 2t = 2 sin t cos t = 2 ⋅ (5/13) ⋅ (-12/13) = -120/169

б) cos 2t

cos 2t = cos² t - sin² t = (-12/13)² - (5/13)² = 144/169 - 25/169 = 119/169

в) tg 2t

tg 2t = \(\frac{sin 2t}{cos 2t}\) = \(\frac{-120/169}{119/169}\) = -120/119

г) ctg 2t

ctg 2t = \(\frac{cos 2t}{sin 2t}\) = \(\frac{119/169}{-120/169}\) = -119/120

3. a) Дано: cos t = 3/4, 0 < t < π/2. Вычислите: cos(t/2); sin(t/2); tg(t/2); ctg(t/2)

cos(t/2) = \(\sqrt{\frac{1 + cos t}{2}}\) = \(\sqrt{\frac{1 + 3/4}{2}}\) = \(\sqrt{\frac{7/4}{2}}\) = \(\sqrt{\frac{7}{8}}\) = \(\frac{\sqrt{14}}{4}\)

sin(t/2) = \(\sqrt{\frac{1 - cos t}{2}}\) = \(\sqrt{\frac{1 - 3/4}{2}}\) = \(\sqrt{\frac{1/4}{2}}\) = \(\sqrt{\frac{1}{8}}\) = \(\frac{\sqrt{2}}{4}\)

tg(t/2) = \(\frac{sin(t/2)}{cos(t/2)}\) = \(\frac{\frac{\sqrt{2}}{4}}{\frac{\sqrt{14}}{4}}\) = \(\frac{\sqrt{2}}{\sqrt{14}}\) = \(\frac{1}{\sqrt{7}}\) = \(\frac{\sqrt{7}}{7}\)

ctg(t/2) = \(\frac{cos(t/2)}{sin(t/2)}\) = \(\frac{\frac{\sqrt{14}}{4}}{\frac{\sqrt{2}}{4}}\) = \(\frac{\sqrt{14}}{\sqrt{2}}\) = \(\sqrt{7}\)

б) Дано: ctg t = 3/4, π < t < 3π/2. Вычислите: cos(t/2); sin(t/2); tg(t/2); ctg(t/2)

tg t = 4/3

sin t = -\(\sqrt{\frac{tg^2 t}{1 + tg^2 t}}\) = -\(\sqrt{\frac{(4/3)^2}{1 + (4/3)^2}}\) = -\(\sqrt{\frac{16/9}{1 + 16/9}}\) = -\(\sqrt{\frac{16/9}{25/9}}\) = -\(\sqrt{\frac{16}{25}}\) = -4/5

cos t = -\(\sqrt{1 - sin^2 t}\) = -\(\sqrt{1 - (-4/5)^2}\) = -\(\sqrt{1 - 16/25}\) = -\(\sqrt{9/25}\) = -3/5

cos(t/2) = -\(\sqrt{\frac{1 + cos t}{2}}\) = -\(\sqrt{\frac{1 - 3/5}{2}}\) = -\(\sqrt{\frac{2/5}{2}}\) = -\(\sqrt{\frac{1}{5}}\) = -\(\frac{\sqrt{5}}{5}\)

sin(t/2) = \(\sqrt{\frac{1 - cos t}{2}}\) = \(\sqrt{\frac{1 + 3/5}{2}}\) = \(\sqrt{\frac{8/5}{2}}\) = \(\sqrt{\frac{4}{5}}\) = \(\frac{2\sqrt{5}}{5}\)

tg(t/2) = \(\frac{sin(t/2)}{cos(t/2)}\) = \(\frac{\frac{2\sqrt{5}}{5}}{-\frac{\sqrt{5}}{5}}\) = -2

ctg(t/2) = \(\frac{cos(t/2)}{sin(t/2)}\) = \(\frac{-\frac{\sqrt{5}}{5}}{\frac{2\sqrt{5}}{5}}\) = -1/2

4. Упростите выражение:

a) \(\frac{sin t}{2 cos^2 (t/2)}\)

\(\frac{sin t}{2 cos^2 (t/2)}\) = \(\frac{2 sin (t/2) cos (t/2)}{2 cos^2 (t/2)}\) = \(\frac{sin (t/2)}{cos (t/2)}\) = tg (t/2)

б) \(\frac{cos t}{cos (t/2) + sin (t/2)}\)

\(\frac{cos t}{cos (t/2) + sin (t/2)}\) = \(\frac{cos^2(t/2)-sin^2(t/2)}{cos(t/2)+sin(t/2)}\)= \(\frac{(cos(t/2) - sin(t/2))(cos(t/2)+sin(t/2))}{cos (t/2) + sin (t/2)}\) = cos(t/2) - sin(t/2)

Ответ: Решения выше