30. Гузориши универсалӣ - Алгебра — синфи 10
Усули гузориши универсалӣ — ин тарзест, ки муодилаи тригонометриро бо як гузориш ба муодилаи ратсионалӣ меорад. Номаълум tg x 2 \operatorname{tg}\dfrac{x}{2} tg 2 x t t t sin x , \sin x, sin x , cos x \cos x cos x
навбатӣ 31. Системаи муодилаҳои тригонометрӣ 1 соат
t
sin x = 2 t 1 + t 2 , cos x = 1 − t 2 1 + t 2 , t = tg x 2 . \sin x = \frac{2t}{1 + t^2},\qquad \cos x = \frac{1 - t^2}{1 + t^2},\qquad t = \operatorname{tg}\frac{x}{2}. sin x = 1 + t 2 2 t , cos x = 1 + t 2 1 − t 2 , t = tg 2 x . Мисоли 1. sin x + cos x = 1 \sin x + \cos x = 1 sin x + cos x = 1
Ҳал. Бо t = tg x 2 t = \operatorname{tg}\dfrac{x}{2} t = tg 2 x 2 t 1 + t 2 + 1 − t 2 1 + t 2 = 1. \frac{2t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} = 1. 1 + t 2 2 t + 1 + t 2 1 − t 2 = 1. 1 + t 2 ≠ 0 , 1 + t^2 \ne 0, 1 + t 2 = 0 , 1 + t 2 1 + t^2 1 + t 2 2 t + 1 − t 2 = 1 + t 2 , 2t + 1 - t^2 = 1 + t^2, 2 t + 1 − t 2 = 1 + t 2 , 2 t 2 − 2 t = 0 , 2t^2 - 2t = 0, 2 t 2 − 2 t = 0 , 2 t ( t − 1 ) = 0 , 2t(t - 1) = 0, 2 t ( t − 1 ) = 0 , t 1 = 0 , t_1 = 0, t 1 = 0 , t 2 = 1. t_2 = 1. t 2 = 1. tg x 2 = 0 , x 2 = π n , x = 2 π n ; tg x 2 = 1 , x 2 = π 4 + π n , x = π 2 + 2 π n . \operatorname{tg}\frac{x}{2} = 0,\ \frac{x}{2} = \pi n,\ x = 2\pi n;\qquad \operatorname{tg}\frac{x}{2} = 1,\ \frac{x}{2} = \frac{\pi}{4} + \pi n,\ x = \frac{\pi}{2} + 2\pi n. tg 2 x = 0 , x = π x = \pi x = π sin π + cos π = − 1 ≠ 1 \sin\pi + \cos\pi = -1 \ne 1 sin π + cos π = − 1 = 1 x = 2 π n x = 2\pi n x = 2 πn x = π 2 + 2 π n , x = \dfrac{\pi}{2} + 2\pi n, x = 2 π + 2 πn , n ∈ Z . n \in \mathbb{Z}. n ∈ Z .
Мисоли 2. 3 sin x + 4 cos x = 4 3\sin x + 4\cos x = 4 3 sin x + 4 cos x = 4
Ҳал. t = tg x 2 : t = \operatorname{tg}\dfrac{x}{2}: t = tg 2 x : 6 t 1 + t 2 + 4 ( 1 − t 2 ) 1 + t 2 = 4 , \dfrac{6t}{1 + t^2} + \dfrac{4(1 - t^2)}{1 + t^2} = 4, 1 + t 2 6 t + 1 + t 2 4 ( 1 − t 2 ) = 4 , 6 t + 4 − 4 t 2 = 4 + 4 t 2 , 6t + 4 - 4t^2 = 4 + 4t^2, 6 t + 4 − 4 t 2 = 4 + 4 t 2 , 8 t 2 − 6 t = 0 , 8t^2 - 6t = 0, 8 t 2 − 6 t = 0 , 2 t ( 4 t − 3 ) = 0 , 2t(4t - 3) = 0, 2 t ( 4 t − 3 ) = 0 , t 1 = 0 , t_1 = 0, t 1 = 0 , t 2 = 3 4 . t_2 = \dfrac{3}{4}. t 2 = 4 3 . x = 2 π n ; x 2 = arctg 3 4 + π n , x = 2 arctg 3 4 + 2 π n , n ∈ Z . x = 2\pi n;\qquad \frac{x}{2} = \operatorname{arctg}\frac{3}{4} + \pi n,\ x = 2\operatorname{arctg}\frac{3}{4} + 2\pi n,\quad n \in \mathbb{Z}. x = 2 πn ; 2 x = arctg x = π x = \pi x = π 3 ⋅ 0 + 4 ⋅ ( − 1 ) = − 4 ≠ 4 3 \cdot 0 + 4 \cdot (-1) = -4 \ne 4 3 ⋅ 0 + 4 ⋅ ( − 1 ) = − 4 = 4
Мисоли 3. 4 sin x + 3 cos x = − 3 4\sin x + 3\cos x = -3 4 sin x + 3 cos x = − 3
Ҳал. t = tg x 2 : t = \operatorname{tg}\dfrac{x}{2}: t = tg 2 x : 8 t + 3 ( 1 − t 2 ) + 3 ( 1 + t 2 ) = 0 , 8t + 3(1 - t^2) + 3(1 + t^2) = 0, 8 t + 3 ( 1 − t 2 ) + 3 ( 1 + t 2 ) = 0 , 8 t + 6 = 0 , 8t + 6 = 0, 8 t + 6 = 0 , t = − 3 4 . t = -\dfrac{3}{4}. t = − 4 3 . x 2 = − arctg 3 4 + π n , x = − 2 arctg 3 4 + 2 π n . \frac{x}{2} = -\operatorname{arctg}\frac{3}{4} + \pi n,\qquad x = -2\operatorname{arctg}\frac{3}{4} + 2\pi n. 2 x = − arctg 4 3 x = π x = \pi x = π 4 ⋅ 0 + 3 ⋅ ( − 1 ) = − 3 4 \cdot 0 + 3 \cdot (-1) = -3 4 ⋅ 0 + 3 ⋅ ( − 1 ) = − 3 дуруст аст! Пас x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn x = − 2 arctg 3 4 + 2 π n x = -2\operatorname{arctg}\dfrac{3}{4} + 2\pi n x = − 2 arctg 4 3 + 2 πn x = π + 2 π n , x = \pi + 2\pi n, x = π + 2 πn , n ∈ Z . n \in \mathbb{Z}. n ∈ Z .
Мисоли 4. 3 sin x + cos x = 1 3\sin x + \cos x = 1 3 sin x + cos x = 1
Ҳал. t = tg x 2 : t = \operatorname{tg}\dfrac{x}{2}: t = tg 2 x : 6 t + 1 − t 2 = 1 + t 2 , 6t + 1 - t^2 = 1 + t^2, 6 t + 1 − t 2 = 1 + t 2 , 2 t 2 − 6 t = 0 , 2t^2 - 6t = 0, 2 t 2 − 6 t = 0 , t 1 = 0 , t_1 = 0, t 1 = 0 , t 2 = 3. t_2 = 3. t 2 = 3. x = 2 π n x = 2\pi n x = 2 πn x = 2 arctg 3 + 2 π n , x = 2\operatorname{arctg} 3 + 2\pi n, x = 2 arctg 3 + 2 πn , n ∈ Z . n \in \mathbb{Z}. n ∈ Z . x = π x = \pi x = π 0 − 1 = − 1 ≠ 1 0 - 1 = -1 \ne 1 0 − 1 = − 1 = 1
Мисоли 5. sin x + cos x = − 1 \sin x + \cos x = -1 sin x + cos x = − 1
Ҳал. t = tg x 2 : t = \operatorname{tg}\dfrac{x}{2}: t = tg 2 x : 2 t + 1 − t 2 = − 1 − t 2 , 2t + 1 - t^2 = -1 - t^2, 2 t + 1 − t 2 = − 1 − t 2 , 2 t = − 2 , 2t = -2, 2 t = − 2 , t = − 1 , t = -1, t = − 1 , x 2 = − π 4 + π n , \dfrac{x}{2} = -\dfrac{\pi}{4} + \pi n, 2 x = − 4 π + πn , x = − π 2 + 2 π n . x = -\dfrac{\pi}{2} + 2\pi n. x = − 2 π + 2 πn . x = π x = \pi x = π sin π + cos π = − 1 \sin\pi + \cos\pi = -1 sin π + cos π = − 1 x = − π 2 + 2 π n , x = π + 2 π n , n ∈ Z . x = -\frac{\pi}{2} + 2\pi n,\qquad x = \pi + 2\pi n,\quad n \in \mathbb{Z}. x = − 2 π + 2 πn , x = π + 2 πn ,
Ҳангоми кунҷи дучанда (sin 2 x , \sin 2x, sin 2 x , cos 2 x \cos 2x cos 2 x t = tg x t = \operatorname{tg} x t = tg x x = π 2 + π n x = \dfrac{\pi}{2} + \pi n x = 2 π + πn
Кор дар синф
Машқҳо
Дар синф
а) 4 sin x + 5 cos x = 6 ; 4\sin x + 5\cos x = 6; 4 sin x + 5 cos x = 6 ;
б) 3 sin x − 2 cos x = 1 ; \sqrt{3}\sin x - 2\cos x = 1; 3 sin x − 2 cos x = 1 ;
в) 2 ( cos x + sin x ) = 2 ; 2(\cos x + \sin x) = \sqrt{2}; 2 ( cos x + sin x ) = 2 ;
г) sin 2 x + 3 cos 2 x = 2 ; \sin 2x + \sqrt{3}\cos 2x = \sqrt{2}; sin 2 x + 3 cos 2 x = 2
д) sin x − 5 cos x = 5 ; \sin x - \sqrt{5}\cos x = \sqrt{5}; sin x − 5 cos x = 5
е) sin 2 x + cos 2 x = 1. \sin 2x + \cos 2x = 1. sin 2 x + cos 2 x = 1.
Нишондод: дар г) ва е) t = tg x t = \operatorname{tg} x t = tg x x = π 2 + π n x = \dfrac{\pi}{2} + \pi n x = 2 π + πn
Дар хона 2. Системаи нобаробариҳоро ҳал кунед:
а) { 3 x − 18 > 0 , 4 x − 12 > 0 ; \begin{cases} 3x - 18 > 0, \\ 4x - 12 > 0; \end{cases} { 3 x − 18 > 0 , 4 x − 12 > 0 ;
б) { 2 x + 4 ≤ 0 , 4 − 3 x > 0 ; \begin{cases} 2x + 4 \le 0, \\ 4 - 3x > 0; \end{cases} { 2 x + 4 ≤ 0 , 4 − 3 x > 0 ;
в) { 4 x + 2 > 5 x + 3 , 2 − 3 x < 7 − 2 x . \begin{cases} 4x + 2 > 5x + 3, \\ 2 - 3x < 7 - 2x. \end{cases} { 4 x + 2 > 5 x + 3 , 2 − 3 x < 7 − 2 x .
3. Дар байни ададҳои 1 1 1 16 16 16
4. Суммаи n n n 2 3 ; 1 2 ; 3 8 ; … \dfrac{2}{3};\ \dfrac{1}{2};\ \dfrac{3}{8};\ \ldots 3 2 ; 2 1 ; 8 3 ; …
5. Дар прогрессияи геометрӣ b 1 = 1 ; b_1 = 1; b 1 = 1 ; b 3 + b 5 = 90 b_3 + b_5 = 90 b 3 + b 5 = 90
6. Суммаи прогрессияи геометрии беохирро ёбед: 3 2 + 2 3 + 2 3 2 3 + … \sqrt{\dfrac{3}{2}} + \sqrt{\dfrac{2}{3}} + \dfrac{2}{3}\sqrt{\dfrac{2}{3}} + \ldots 2 3 + 3 2 + 3 2 3 2 + …
Вазифаи хонагӣ
Машқҳои 2–6; аз машқи 1 ҳал нашудаҳо.
Вазифаи хонагии онлайн: «Алгебра 10 · Вазифаи хонагӣ: 30. Гузориши универсалӣ».
Соати оянда: 31 .
Хулоса
t = tg x 2 t = \operatorname{tg}\dfrac{x}{2} t = tg 2 x sin x = 2 t 1 + t 2 , \sin x = \dfrac{2t}{1 + t^2}, sin x = 1 + t 2 2 t , cos x = 1 − t 2 1 + t 2 . \cos x = \dfrac{1 - t^2}{1 + t^2}. cos x = 1 + t 2 1 − t 2 . a sin x + b cos x = c a\sin x + b\cos x = c a sin x + b cos x = c t t t Гузориш ҳамеша решаи x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn
Барои кунҷи дучанда гузориш t = tg x t = \operatorname{tg} x t = tg x x = π 2 + π n . x = \dfrac{\pi}{2} + \pi n. x = 2 π + πn .
Худро санҷед. Ҷавоб диҳед «дуруст» ё «нодуруст»:
Гузориши универсалӣ sin x \sin x sin x cos x \cos x cos x tg x 2 \operatorname{tg}\dfrac{x}{2} tg 2 x
Гузориши универсалӣ ҳеҷ гоҳ решаро гум намекунад. ✗
Барои 4 sin x + 3 cos x = − 3 4\sin x + 3\cos x = -3 4 sin x + 3 cos x = − 3 x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn
Бо t = tg x 2 t = \operatorname{tg}\dfrac{x}{2} t = tg 2 x
Дар ин дарс
Ман медонам формулаҳои sin x \sin x sin x cos x \cos x cos x t = tg x 2 . t = \operatorname{tg}\dfrac{x}{2}. t = tg 2 x . Ман метавонам муодилаи a sin x + b cos x = c a\sin x + b\cos x = c a sin x + b cos x = c Ман метавонам ҳолати x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn Такрор (2 дақиқа)
sin x \sin x sin x cos x \cos x cos x tg x 2 \operatorname{tg}\dfrac{x}{2} tg 2 x tg x 2 \operatorname{tg}\dfrac{x}{2} tg 2 x x x x sin π \sin\pi sin π cos π \cos\pi cos π Ҷавобҳо
sin x = 2 t 1 + t 2 , \sin x = \dfrac{2t}{1 + t^2}, sin x = 1 + t 2 2 t , cos x = 1 − t 2 1 + t 2 , \cos x = \dfrac{1 - t^2}{1 + t^2}, cos Вазъият барои андешаронӣ Муодилаи sin x + cos x = 1 \sin x + \cos x = 1 sin x + cos x = 1 t t t
Дар хотир доред Формулаҳо танҳо барои x ≠ π + 2 π n x \ne \pi + 2\pi n x = π + 2 πn x = π + 2 π n x = \pi + 2\pi n x = π +
Калимаҳои калидӣ гузориши универсалӣ тангенси нимкунҷ ифодаи ратсионалӣ решаи гумшуда санҷиш
Ман дар ин дарс чӣ омӯхтам? Чаро x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn
Гузориши универсалӣ t = tg x 2 t = \operatorname{tg}\dfrac{x}{2} t = tg 2 x x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn
Кадом муодила x = π x = \pi x = π
Як ҷавобро интихоб кунед A 2 sin x + cos x = 1 2\sin x + \cos x = 1 2 sin x + cos x = 1
B 3 sin x − 2 cos x = 2 3\sin x - 2\cos x = 2 3 sin x − 2
C D sin x − cos x = 1 \sin x - \cos x = 1 sin x − cos x = 1
Як ҷавобро интихоб кунед A x = π 2 + 2 π n x = \dfrac{\pi}{2} + 2\pi n x = 2 π + 2 πn
B C D 2
x
=
πn , x =
2 πn ; tg 2 x =
1 , 2 x =
4 π +
πn , x =
2 π +
2 πn .
4
3
+
πn , x =
2 arctg 4 3 +
2 πn , n ∈
Z .
+
πn , x =
− 2 arctg 4 3 +
2 πn .
n
∈
Z .
;
;
x
=
1 + t 2 1 − t 2 ,
t = tg x 2 . t = \operatorname{tg}\dfrac{x}{2}. t = tg 2 x . x 2 = π 2 + π n , \dfrac{x}{2} = \dfrac{\pi}{2} + \pi n, 2 x = 2 π + πn , x = π + 2 π n . x = \pi + 2\pi n. x = π + 2 πn .
2
πn
x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn гум кунад
x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn 4 t 2 − 2 t = 0 4t^2 - 2t = 0 4 t 2 − 2 t = 0
cos
x
=
2
sin x + cos x = 1 \sin x + \cos x = 1 sin x + cos x = 1
sin x − cos x = − 1 \sin x - \cos x = -1 sin x − cos x = − 1
x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn x = π + 2 π n x = \pi + 2\pi n x = π + 2 πn
x = π 2 + 2 π n x = \dfrac{\pi}{2} + 2\pi n x = 2 π + 2 πn
x = π 4 + π n x = \dfrac{\pi}{4} + \pi n x = 4 π + πn