{"id":297766,"date":"2023-12-21T21:16:16","date_gmt":"2023-12-21T14:16:16","guid":{"rendered":"https:\/\/hoanghamobile.com\/tin-tuc\/?p=297766"},"modified":"2025-10-18T19:25:52","modified_gmt":"2025-10-18T12:25:52","slug":"cong-thuc-luong-giac","status":"publish","type":"post","link":"https:\/\/hoanghamobile.com\/tin-tuc\/cong-thuc-luong-giac\/","title":{"rendered":"T\u1ed5ng H\u1ee3p B\u1ea3ng C\u00f4ng Th\u1ee9c L\u01b0\u1ee3ng Gi\u00e1c \u0110\u1ea7y \u0110\u1ee7 & D\u1ec5 Nh\u1edb Nh\u1ea5t"},"content":{"rendered":"

\u1ede m\u1ee9c \u0111\u1ed9 c\u01a1 b\u1ea3n, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 gi\u1ea3i c\u00e1c v\u1ea5n \u0111\u1ec1 h\u00ecnh h\u1ecdc, nh\u01b0 t\u00ednh to\u00e1n chi\u1ec1u d\u00e0i c\u00e1c c\u1ea1nh v\u00e0 \u0111o g\u00f3c trong tam gi\u00e1c vu\u00f4ng. Ngo\u00e0i ra, ch\u00fang c\u00f2n xu\u1ea5t hi\u1ec7n trong nhi\u1ec1u \u1ee9ng d\u1ee5ng th\u1ef1c t\u1ebf t\u1eeb k\u1ef9 thu\u1eadt v\u00e0 v\u1eadt l\u00fd \u0111\u1ebfn m\u00e1y t\u00ednh v\u00e0 ngh\u1ec7 thu\u1eadt \u0111\u1ed3 h\u1ecda. Trong l\u0129nh v\u1ef1c v\u1eadt l\u00fd, t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c gi\u00fap m\u00f4 t\u1ea3 c\u00e1c quy lu\u1eadt \u0111\u1ed1i v\u1edbi c\u00e1c v\u1ea5n \u0111\u1ec1 nh\u01b0 dao \u0111\u1ed9ng v\u00e0 s\u00f3ng. Trong c\u00f4ng ngh\u1ec7, ch\u00fang \u0111\u01b0\u1ee3c t\u00edch h\u1ee3p trong c\u00e1c h\u1ec7 th\u1ed1ng \u0111i\u1ec1u khi\u1ec3n t\u1ef1 \u0111\u1ed9ng, \u0111\u1eb7c bi\u1ec7t l\u00e0 trong vi\u1ec7c \u0111i\u1ec1u khi\u1ec3n v\u1ecb tr\u00ed c\u1ee7a c\u00e1c thi\u1ebft b\u1ecb nh\u01b0 robot. Trong b\u00e0i vi\u1ebft h\u00f4m nay ch\u00fang ta s\u1ebd c\u00f9ng t\u00ecm hi\u1ec3u b\u1ea3ng c\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c \u0111\u1ea7y \u0111\u1ee7 c\u0169ng nh\u01b0 nh\u1eefng \u1ee9ng d\u1ee5ng s\u00e2u r\u1ed9ng c\u1ee7a n\u00f3 trong cu\u1ed9c s\u1ed1ng.<\/span><\/p>\n

Kh\u00e1i ni\u1ec7m t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c<\/b><\/h2>\n

T\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a m\u1ed9t g\u00f3c nh\u1ecdn l\u00e0 m\u1ed9t kh\u00e1i ni\u1ec7m trong l\u01b0\u1ee3ng gi\u00e1c \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 m\u00f4 t\u1ea3 m\u1ed1i quan h\u1ec7 gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c vu\u00f4ng v\u00e0 g\u00f3c t\u01b0\u01a1ng \u1ee9ng. T\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c \u0111\u01b0\u1ee3c \u0111\u1ecbnh ngh\u0129a b\u1eb1ng ph\u00e9p chia c\u1ee7a c\u00e1c c\u1ea1nh trong tam gi\u00e1c vu\u00f4ng. Trong tam gi\u00e1c vu\u00f4ng, c\u00f3 b\u1ed1n t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c ch\u00ednh \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh:<\/span><\/p>\n

1. T\u1ec9 s\u1ed1 sin:<\/b> T\u1ec9 s\u1ed1 sin c\u1ee7a m\u1ed9t g\u00f3c (x) trong tam gi\u00e1c vu\u00f4ng b\u1eb1ng \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u1ed1i c\u1ee7a g\u00f3c chia cho \u0111\u1ed9 d\u00e0i c\u1ea1nh huy\u1ec1n.<\/span><\/p>\n

C\u00f4ng th\u1ee9c: sin(x) = c\u1ea1nh \u0111\u1ed1i\/huy\u1ec1n<\/b><\/p>\n

2. T\u1ec9 s\u1ed1 cos:<\/b> T\u1ec9 s\u1ed1 cos c\u1ee7a m\u1ed9t g\u00f3c (x) trong tam gi\u00e1c vu\u00f4ng b\u1eb1ng \u0111\u1ed9 d\u00e0i c\u1ea1nh k\u1ec1 g\u00f3c chia cho \u0111\u1ed9 d\u00e0i c\u1ea1nh huy\u1ec1n.<\/span><\/p>\n

C\u00f4ng th\u1ee9c: cos(x) = c\u1ea1nh k\u1ec1\/huy\u1ec1n<\/b><\/p>\n

3. T\u1ec9 s\u1ed1 tan:<\/b> T\u1ec9 s\u1ed1 tan c\u1ee7a m\u1ed9t g\u00f3c (x) trong tam gi\u00e1c vu\u00f4ng b\u1eb1ng \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u1ed1i c\u1ee7a g\u00f3c chia cho \u0111\u1ed9 d\u00e0i c\u1ea1nh k\u1ec1 g\u00f3c.<\/span><\/p>\n

C\u00f4ng th\u1ee9c: tan(x) = c\u1ea1nh \u0111\u1ed1i\/c\u1ea1nh k\u1ec1<\/b><\/p>\n

4. T\u1ec9 s\u1ed1 cot:<\/b> T\u1ec9 s\u1ed1 cot c\u1ee7a m\u1ed9t g\u00f3c (x) trong tam gi\u00e1c vu\u00f4ng b\u1eb1ng \u0111\u1ed9 d\u00e0i li\u1ec1n k\u1ec1 chia cho \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u1ed1i c\u1ee7a g\u00f3c.<\/span><\/p>\n

C\u00f4ng th\u1ee9c: cot(x) = c\u1ea1nh k\u1ec1\/c\u1ea1nh \u0111\u1ed1i<\/b><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-1\"<\/picture><\/p>\n

C\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c n\u00e0y \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n c\u00e1c gi\u00e1 tr\u1ecb g\u00f3c, c\u1ea1nh ho\u1eb7c c\u00e1c quan h\u1ec7 gi\u1eefa ch\u00fang trong tam gi\u00e1c vu\u00f4ng. Ch\u00fang c\u0169ng \u0111\u01b0\u1ee3c \u00e1p d\u1ee5ng r\u1ed9ng r\u00e3i trong c\u00e1c l\u0129nh v\u1ef1c nh\u01b0 to\u00e1n h\u1ecdc, v\u1eadt l\u00fd, \u0111\u1ecba h\u00ecnh v\u00e0 k\u1ef9 thu\u1eadt. V\u1ec1 \u1ee9ng d\u1ee5ng c\u1ee7a c\u00e1c t\u1ec9 s\u1ed1 n\u00e0y ch\u00fang ta s\u1ebd t\u00ecm hi\u1ec3u k\u1ef9 h\u01a1n trong n\u1ed9i dung cu\u1ed1i. Trong ph\u1ea7n ti\u1ebfp theo ch\u00fang ta s\u1ebd l\u1ea7n l\u01b0\u1ee3t \u0111i\u1ec3m qua m\u1ed9t lo\u1ea1t c\u00e1c c\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c th\u01b0\u1eddng g\u1eb7p nh\u00e9.<\/span><\/p>\n

C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c c\u01a1 b\u1ea3n<\/b><\/h2>\n

sin^2(x) + cos^2(x) = 1<\/span><\/p>\n

1\/(sin^2(x)) = 1+ cot^2(x)<\/span><\/p>\n

1\/(cos^2(x)) = 1+ tan^2(x)<\/span><\/p>\n

tan(x).cot(x) = 1<\/span><\/p>\n

cot(x) = cos(x)\/sin(x)<\/span><\/p>\n

tan(x) = sin(x)\/cos(x)<\/span><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-13\"<\/picture><\/p>\n

C\u00f4ng th\u1ee9c c\u1ed9ng l\u01b0\u1ee3ng gi\u00e1c<\/b><\/h2>\n

sin(a+b) = sin(a).cos(b) + cos(a).sin(b)<\/span><\/p>\n

sin(a-b) = sin(a).cos(b) – cos(a).sin(b)<\/span><\/p>\n

cos(a+b) = cos(a).cos(b) – sin(a).sin(b)<\/span><\/p>\n

cos(a-b) = cos(a).cos(b) + sin(a).sin(b)<\/span><\/p>\n

tan(a+b) = (tan(a)+tan(b)) \/ (1 – tan(a).tan(b))<\/span><\/p>\n

tan(a-b) = (tan(a)-tan(b)) \/ (1 + tan(a).tan(b))<\/span><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-15\"<\/picture><\/p>\n

C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c bi\u1ebfn \u0111\u1ed5i t\u1ed5ng th\u00e0nh t\u00edch<\/b><\/h2>\n

cos(a) + cos(b) = 2cos((a+b)\/2).cos((a-b)\/2)<\/span><\/p>\n

cos(a) – cos(b) = -2sin((a+b)\/2).sin((a-b)\/2)<\/span><\/p>\n

sin(a) + sin(b) = 2sin((a+b)\/2).cos((a-b)\/2)<\/span><\/p>\n

sin(a) – sin(b) = 2cos((a+b)\/2).sin((a-b)\/2)<\/span><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-14\"<\/picture><\/p>\n

C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c bi\u1ebfn \u0111\u1ed5i t\u00edch th\u00e0nh t\u1ed5ng<\/b><\/h2>\n

cos(a).cos(b) = 1\/2[cos(a+b) + cos(a-b)]<\/span><\/p>\n

sin(a).sin(b) = -1\/2[cos(a+b) – cos(a-b)]<\/span><\/p>\n

sin(a).cos(b) = 1\/2[sin(a+b) + sin(a-b)]<\/span><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-11\"<\/picture><\/p>\n

C\u00e1c c\u00f4ng th\u1ee9c nh\u00e2n g\u00f3c<\/b><\/h2>\n

C\u00e1c c\u00f4ng th\u1ee9c nh\u00e2n l\u01b0\u1ee3ng gi\u00e1c n\u00e0y \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 \u0111\u01a1n gi\u1ea3n h\u00f3a v\u00e0 t\u00ednh to\u00e1n c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c nh\u00e2n 2, nh\u00e2n 3 c\u1ee7a m\u1ed9t g\u00f3c d\u1ef1a tr\u00ean gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a g\u00f3c ban \u0111\u1ea7u.<\/span><\/p>\n

C\u00f4ng th\u1ee9c nh\u00e2n \u0111\u00f4i<\/b><\/h3>\n

C\u00f3 ba c\u00f4ng th\u1ee9c \u0111\u00f4i l\u01b0\u1ee3ng gi\u00e1c nh\u00e2n \u0111\u00f4i t\u01b0\u01a1ng \u1ee9ng v\u1edbi c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c sin, cos v\u00e0 tan:<\/span><\/p>\n

C\u00f4ng th\u1ee9c sin nh\u00e2n \u0111\u00f4i: <\/b>sin(2x) = 2sin(x)cos(x)<\/span><\/p>\n

C\u00f4ng th\u1ee9c cos nh\u00e2n \u0111\u00f4i: <\/b>cos(2x) = cos^2(x) – sin^2(x) = 2cos^2(x) – 1 = 1 – 2sin^2(x)<\/span><\/p>\n

C\u00f4ng th\u1ee9c tan nh\u00e2n \u0111\u00f4i: <\/b>tan(2x) = (2tan(x))\/(1 – tan^2(x))<\/span><\/p>\n

\"cong-thuc-luong-giac-day-du-nhat-18\"<\/picture><\/p>\n

C\u00f4ng th\u1ee9c nh\u00e2n ba<\/b><\/h3>\n

C\u00f3 hai c\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c nh\u00e2n ba ch\u00ednh, t\u01b0\u01a1ng \u1ee9ng v\u1edbi c\u00e1c t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c sin v\u00e0 cos:<\/span><\/p>\n

C\u00f4ng th\u1ee9c sin nh\u00e2n ba: <\/b>sin(3x) = 3sin(x) – 4sin^3(x)<\/span><\/p>\n

C\u00f4ng th\u1ee9c cos nh\u00e2n ba: <\/b>cos(3x) = 4cos^3(x) – 3cos(x)<\/span><\/p>\n

C\u00f4ng th\u1ee9c tan nh\u00e2n ba: <\/b>tan(3x) = (3tan(x) – tan^3(x))\/(1 – 3tan^2(x))<\/span><\/p>\n

Ph\u00e9p t\u00ednh l\u01b0\u1ee3ng gi\u00e1c c\u00e1c cung \u0111\u1eb7c bi\u1ec7t<\/b><\/h2>\n

C\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c cung \u0111\u1eb7c bi\u1ec7t l\u00e0 c\u00e1c ph\u00e9p t\u00ednh l\u01b0\u1ee3ng gi\u00e1c \u00e1p d\u1ee5ng cho c\u00e1c g\u00f3c \u0111\u1eb7c bi\u1ec7t nh\u01b0 g\u00f3c 0 \u0111\u1ed9, 90 \u0111\u1ed9 v\u00e0 180 \u0111\u1ed9. C\u00e1c ph\u00e9p t\u00ednh l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c cung \u0111\u1eb7c bi\u1ec7t gi\u00fap t\u00ednh to\u00e1n nhanh ch\u00f3ng c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c g\u00f3c c\u1ee5 th\u1ec3 m\u00e0 kh\u00f4ng c\u1ea7n d\u00f9ng \u0111\u1ebfn b\u1ea3ng gi\u00e1 tr\u1ecb ho\u1eb7c m\u00e1y t\u00ednh. \u0110i\u1ec1u n\u00e0y ti\u1ebft ki\u1ec7m th\u1eddi gian v\u00e0 gi\u00fap t\u0103ng t\u1ed1c \u0111\u1ed9 gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n l\u01b0\u1ee3ng gi\u00e1c. C\u00f3 ba ph\u00e9p t\u00ednh l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c cung li\u00ean quan \u0111\u1eb7c bi\u1ec7t, \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 t\u00ednh to\u00e1n c\u00e1c gi\u00e1 tr\u1ecb l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a c\u00e1c cung g\u00f3c \u0111\u1eb7c bi\u1ec7t.<\/span><\/p>\n

Ph\u00e9p t\u00ednh l\u01b0\u1ee3ng gi\u00e1c c\u1ee7a cung n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n:<\/b><\/p>\n