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{"id":44365,"date":"2019-10-16T16:24:31","date_gmt":"2019-10-16T09:24:31","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44365"},"modified":"2019-10-16T16:24:36","modified_gmt":"2019-10-16T09:24:36","slug":"giai-toan-10-chuong-6-cung-va-goc-luong-giac-cong-thuc-luong-giac-bai-2-gia-tri-luong-giac-cua-mot-cung","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-6-cung-va-goc-luong-giac-cong-thuc-luong-giac-bai-2-gia-tri-luong-giac-cua-mot-cung\/","title":{"rendered":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 6: Cung v\u00e0 g\u00f3c l\u01b0\u1ee3ng gi\u00e1c – C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c\/ B\u00e0i 2: Gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a m\u1ed9t cung"},"content":{"rendered":"\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 141<\/strong>: Nh\u1eafc l\u1ea1i kh\u00e1i ni\u1ec7m gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c \u03b1, 0o<\/sup> \u2264 \u03b1 \u2264 180o<\/sup>.<\/ins><\/p>\n\n\n\n

Ta c\u00f3 th\u1ec3 m\u1edf r\u1ed9ng kh\u00e1i ni\u1ec7m gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c cho c\u00e1c cung v\u00e0 g\u00f3c l\u01b0\u1ee3ng gi\u00e1c.<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

C\u00e1c s\u1ed1 sin\u2061\u03b1; cos\u2061\u03b1; tan\u2061\u03b1; cot\u2061\u03b1 \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c \u03b1, v\u1edbi 0o<\/sup> \u2264 \u03b1 \u2264 180o<\/sup>.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 142<\/strong>: T\u00ednh sin 25\u03c0\/4, cos(-240o<\/sup>), tan(-405o<\/sup>).<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

sin 25\u03c0\/4 = sin(6\u03c0 + \u03c0\/4) = sin \u03c0\/4 = \u221a2\/2<\/p>\n\n\n\n

cos(-240o<\/sup> ) = cos\u2061(-180o<\/sup> – 60o<\/sup>) = cos\u2061(-60o<\/sup>) = cos\u206160o<\/sup> = 1\/2 <\/p>\n\n\n\n

tan\u2061(-405o<\/sup> ) = tan\u2061(-360o<\/sup> – 45o<\/sup>) = -tan\u206145o<\/sup> = -1 <\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 143<\/strong>: T\u1eeb \u0111\u1ecbnh ngh\u0129a c\u1ee7a sin\uf061 v\u00e0 cos\uf061, h\u00e3y ph\u00e1t bi\u1ec3u \u00fd ngh\u0129a h\u00ecnh h\u1ecdc c\u1ee7a ch\u00fang.<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

sin\u03b1 \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n b\u1edfi \u0111\u1ed9 d\u00e0i \u0111\u1ea1i s\u1ed1 c\u1ee7a vecto (OK) tr\u00ean tr\u1ee5c Oy. Tr\u1ee5c Oy l\u00e0 tr\u1ee5c sin.<\/p>\n\n\n\n

cos\u03b1 \u0111\u01b0\u1ee3c bi\u1ec3u di\u1ec5n b\u1edfi \u0111\u1ed9 d\u00e0i \u0111\u1ea1i s\u1ed1 c\u1ee7a vecto (OH) tr\u00ean tr\u1ee5c Ox. Tr\u1ee5c Oy l\u00e0 tr\u1ee5c cos.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 145<\/strong>: T\u1eeb \u00fd ngh\u0129a h\u00ecnh h\u1ecdc c\u1ee7a tan\u03b1 v\u00e0 cot\u03b1 h\u00e3y suy ra v\u1edbi m\u1ecdi s\u1ed1 nguy\u00ean k, tan(\u03b1 + k\u03c0) = tan\u03b1, cot(\u03b1 + k\u03c0) = cot\u03b1.<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

Khi \u03b2 = \u03b1 + k\u03c0 th\u00ec \u0111i\u1ec3m cu\u1ed1i c\u1ee7a g\u00f3c \u03b2 s\u1ebd tr\u00f9ng v\u1edbi \u0111i\u1ec3m T tr\u00ean tr\u1ee5c tan. Do \u0111\u00f3 <\/p>\n\n\n\n

tan(\u03b1 + k\u03c0) = tan\u03b1.<\/p>\n\n\n\n

Khi \u03b2 = \u03b1 + k\u03c0 th\u00ec \u0111i\u1ec3m cu\u1ed1i c\u1ee7a g\u00f3c \u03b2 s\u1ebd tr\u00f9ng v\u1edbi \u0111i\u1ec3m S tr\u00ean tr\u1ee5c cot. Do \u0111\u00f3 <\/p>\n\n\n\n

cot(\u03b1 + k\u03c0) = cot\u03b1.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 145<\/strong>: T\u1eeb \u0111\u1ecbnh ngh\u0129a c\u1ee7a sin\u03b1, cos\u03b1. H\u00e3y ch\u1ee9ng minh h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ea7u ti\u00ean, t\u1eeb \u0111\u00f3 suy ra c\u00e1c h\u1eb1ng \u0111\u1eb3ng th\u1ee9c c\u00f2n l\u1ea1i.<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

sin\u03b1 = (OK) ;cos\u03b1 = (OH)<\/p>\n\n\n\n

Do tam gi\u00e1c OMK vu\u00f4ng t\u1ea1i K n\u00ean: <\/p>\n\n\n\n

sin2<\/sup> \u03b1 + cos2<\/sup> \u03b1 = OK2<\/sup> + OH2<\/sup> = OK2<\/sup> + MK2<\/sup> = OM2<\/sup> = 1.<\/p>\n\n\n\n

V\u1eady sin2<\/sup> \u03b1 + cos2<\/sup> \u03b1 = 1.<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 148<\/strong>: T\u00ednh cos(-11\u03c0\/4), tan31\u03c0\/6, sin(-1380o<\/sup>).<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 1 (trang 148 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: C\u00f3 cung \u03b1 n\u00e0o m\u00e0 sin\u03b1 nh\u1eadn c\u00e1c gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng sau \u0111\u00e2y kh\u00f4ng ?<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

Ta c\u00f3: -1 \u2264 sin \u03b1 \u2264 1 v\u1edbi m\u1ecdi \u03b1 \u2208 R. <\/p>\n\n\n\n

a) V\u00ec -1 < \u20130,7 < 1 n\u00ean t\u1ed3n t\u1ea1i cung \u03b1 th\u1ecfa m\u00e3n sin \u03b1 = -0,7. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Tr\u00ean tr\u1ee5c tung x\u00e1c \u0111\u1ecbnh \u0111i\u1ec3m K sao cho \n<\/p>\n\n\n\n

T\u1eeb K k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi tr\u1ee5c tung c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n l\u01b0\u1ee3ng gi\u00e1c t\u1ea1i hai \u0111i\u1ec3m M1<\/sub> v\u00e0 M2<\/sub>. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Khi \u0111\u00f3 v\u1edbi \n th\u00ec theo \u0111\u1ecbnh ngh\u0129a sin \u03b1 = \n<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

b) V\u00ec 4\/3 > 1 n\u00ean kh\u00f4ng t\u1ed3n t\u1ea1i \u03b1 \u0111\u1ec3 sin \u03b1 = 4\/3. <\/p>\n\n\n\n

c) V\u00ec -\u221a2 < -1 n\u00ean kh\u00f4ng t\u1ed3n t\u1ea1i \u03b1 \u0111\u1ec3 sin \u03b1 = -\u221a2. <\/p>\n\n\n\n

d) V\u00ec \u221a5\/2 > 1 n\u00ean kh\u00f4ng t\u1ed3n t\u1ea1i \u03b1 \u0111\u1ec3 sin \u03b1 = \u221a5\/2<\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

+ \u0110\u1ecbnh ngh\u0129a sin c\u1ee7a cung \u03b1: <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Cung c\u00f3 s\u0111 = \u03b1. K l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m M tr\u00ean tr\u1ee5c tung. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Khi \u0111\u00f3 ta \u0111\u1ecbnh ngh\u0129a <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

( l\u00e0 \u0111\u1ed9 d\u00e0i \u0111\u1ea1i s\u1ed1 c\u1ee7a OK v\u00e0 ch\u00ednh l\u00e0 tung \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m M).<\/p>\n\n\n\n

+ V\u1edbi m\u1ecdi \u03b1 \u2208 R th\u00ec -1 \u2264 sin \u03b1 \u2264 1. <\/p>\n\n\n\n

B\u00e0i 2 (trang 148 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: C\u00e1c \u0111\u1eb3ng th\u1ee9c sau \u0111\u00e2y c\u00f3 th\u1ec3 \u0111\u1ed3ng th\u1eddi x\u1ea3y ra kh\u00f4ng ?<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 3 (trang 148 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Cho 0 < \u03b1 < \u03c0\/2. X\u00e1c \u0111\u1ecbnh d\u1ea5u c\u1ee7a c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

V\u00ec 0 < \u03b1 < \u03c0\/2 n\u00ean sin \u03b1 > 0, cos \u03b1 > 0, tan \u03b1 > 0, cot \u03b1 > 0. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

C\u00e1ch 1: D\u1ef1a v\u00e0o m\u1ed1i quan h\u1ec7 gi\u1eefa c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c cung c\u00f3 li\u00ean quan \u0111\u1eb7c bi\u1ec7t <\/p>\n\n\n\n

a) sin (\u03b1 \u2013 \u03c0) = – sin (\u03c0 \u2013 \u03b1) \t(\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c sin (- \u03b1) = – sin \u03b1)<\/p>\n\n\n\n

= -sin \u03b1 \t\t(\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c sin (\u03c0 \u2013 \u03b1) = sin \u03b1)<\/p>\n\n\n\n

M\u00e0 sin \u03b1 > 0 n\u00ean sin (\u03b1 \u2013 \u03c0) < 0. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

c) tan (\u03b1 + \u03c0) = tan \u03b1. <\/p>\n\n\n\n

M\u00e0 tan \u03b1 > 0 n\u00ean tan (\u03b1 + \u03c0) > 0.<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

C\u00e1ch 2: D\u1ef1a v\u00e0o bi\u1ec3u di\u1ec5n cung tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n l\u01b0\u1ee3ng gi\u00e1c: <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u00ec 0 < \u03b1 < \u03c0\/2 n\u00ean ta bi\u1ec3u di\u1ec5n \u03b1 = s\u0111 \n nh\u01b0 tr\u00ean h\u00ecnh v\u1ebd. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 4 (trang 148 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: T\u00ednh c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c \u03b1 n\u1ebfu<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 5 (trang 148 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: T\u00ednh \u03b1, bi\u1ebft <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

a) cos \u03b1 = 1 \u21d4 M tr\u00f9ng v\u1edbi A hay \u03b1 = k.2\u03c0, k \u2208 Z.<\/p>\n\n\n\n

b) cos \u03b1 = -1 \u21d4 M tr\u00f9ng v\u1edbi A\u2019 hay \u03b1 = \u03c0 + k.2\u03c0, k \u2208 Z<\/p>\n\n\n\n

c) cos \u03b1 = 0 \u21d4 M tr\u00f9ng v\u1edbi B ho\u1eb7c B\u2019 hay \u03b1 = \u03c0\/2 + k.\u03c0, k \u2208 Z<\/p>\n\n\n\n

d) sin \u03b1 = 1 \u21d4 M tr\u00f9ng v\u1edbi B hay \u03b1 = \u03c0\/2 + k.2\u03c0, k \u2208 Z<\/p>\n\n\n\n

e) sin \u03b1 = -1 \u21d4 M tr\u00f9ng v\u1edbi B\u2019 hay \u03b1 = -\u03c0\/2 + k.2\u03c0, k \u2208 Z<\/p>\n\n\n\n

f) sin \u03b1 = 0 \u21d4 M tr\u00f9ng v\u1edbi A ho\u1eb7c A\u2019 hay \u03b1 = k.\u03c0, k \u2208 Z<\/ins><\/p>\n","protected":false},"excerpt":{"rendered":"

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 141: Nh\u1eafc l\u1ea1i kh\u00e1i ni\u1ec7m gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c \u03b1, 0o \u2264 \u03b1 \u2264 180o. Ta c\u00f3 th\u1ec3 m\u1edf r\u1ed9ng kh\u00e1i ni\u1ec7m gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c cho c\u00e1c cung v\u00e0 g\u00f3c l\u01b0\u1ee3ng gi\u00e1c. L\u1eddi gi\u1ea3i C\u00e1c s\u1ed1 sin\u2061\u03b1; cos\u2061\u03b1; tan\u2061\u03b1; cot\u2061\u03b1 \u0111\u01b0\u1ee3c […]<\/p>\n","protected":false},"author":12,"featured_media":44366,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 6: Cung v\u00e0 g\u00f3c l\u01b0\u1ee3ng gi\u00e1c - C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c\/ B\u00e0i 2: Gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a m\u1ed9t cung<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-6-cung-va-goc-luong-giac-cong-thuc-luong-giac-bai-2-gia-tri-luong-giac-cua-mot-cung\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 6: Cung v\u00e0 g\u00f3c l\u01b0\u1ee3ng gi\u00e1c - C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c\/ B\u00e0i 2: Gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a m\u1ed9t cung\" \/>\n<meta property=\"og:description\" content=\"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 141: Nh\u1eafc l\u1ea1i kh\u00e1i ni\u1ec7m gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c \u03b1, 0o \u2264 \u03b1 \u2264 180o. 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