L\u01b0\u1ee3ng gi\u00e1c l\u00e0 d\u1ea1ng b\u00e0i xu\u1ea5t hi\u1ec7n trong \u0111\u1ec1 thi m\u00f4n to\u00e1n THPT Qu\u1ed1c gia nh\u01b0ng thay v\u00ec nh\u1edb nh\u1eefng c\u00f4ng th\u1ee9c kh\u00f4 khan n\u00e0y, t\u1ea1i sao b\u1ea1n kh\u00f4ng th\u1eed bi\u1ebfn n\u00f3 th\u00e0nh nh\u1eefng c\u00e2u th\u01a1 nh\u1ec9.<\/p>\n

H\u00c0M S\u1ed0 L\u01af\u1ee2NG GI\u00c1C
\n<\/strong>
\nB\u1eaft \u0111\u01b0\u1ee3c qu\u1ea3 tang
\nSin n\u1eb1m tr\u00ean cos (tan@ = sin@:cos@)
\nCotang d\u1ea1i d\u1ed9t
\nB\u1ecb cos \u0111\u00e8 cho. (cot@ = cos@:sin@)
\nVersion 2:
\nB\u1eaft \u0111\u01b0\u1ee3c qu\u1ea3 tang
\nSin n\u1eb1m tr\u00ean cos
\nC\u00f4tang c\u00e3i l\u1ea1i
\nCos n\u1eb1m tr\u00ean sin!<\/p>\n

GI\u00c1 TR\u1eca L\u01af\u1ee2NG GI\u00c1C C\u1ee6A C\u00c1C CUNG \u0110\u1eb6C BI\u1ec6T<\/strong><\/p>\n

Cos \u0111\u1ed1i, sin b\u00f9, ph\u1ee5 ch\u00e9o, kh\u00e1c pi tan<\/p>\n

(Cosin c\u1ee7a hai g\u00f3c \u0111\u1ed1i b\u1eb1ng nhau; sin c\u1ee7a hai g\u00f3c b\u00f9 nhau th\u00ec b\u1eb1ng nhau; ph\u1ee5 ch\u00e9o l\u00e0 2 g\u00f3c ph\u1ee5 nhau th\u00ec sin g\u00f3c n\u00e0y = cos g\u00f3c kia, tan g\u00f3c n\u00e0y = cot g\u00f3c kia; tan c\u1ee7a hai g\u00f3c h\u01a1n k\u00e9m pi th\u00ec b\u1eb1ng nhau)<\/p>\n

C\u00d4NG TH\u1ee8C C\u1ed8NG<\/strong><\/p>\n

Cos c\u1ed9ng cos b\u1eb1ng hai cos cos
\ncos tr\u1eeb cos b\u1eb1ng tr\u1eeb hai sin sin
\nSin c\u1ed9ng sin b\u1eb1ng hai sin cos
\nsin tr\u1eeb sin b\u1eb1ng hai cos sin.<\/p>\n

Sin th\u00ec sin cos cos sin
\nCos th\u00ec cos cos sin sin \u201ccoi ch\u1eebng\u201d (d\u1ea5u tr\u1eeb).
\nTang t\u1ed5ng th\u00ec l\u1ea5y t\u1ed5ng tang
\nChia m\u1ed9t tr\u1eeb v\u1edbi t\u00edch tang, d\u1ec5 \u00f2m.<\/p>\n

C\u00d4NG TH\u1ee8C NH\u00c2N BA<\/strong><\/p>\n

Nh\u00e2n ba m\u1ed9t g\u00f3c b\u1ea5t k\u1ef3,
\nsin th\u00ec ba b\u1ed1n, cos th\u00ec b\u1ed1n ba,
\nd\u1ea5u tr\u1eeb \u0111\u1eb7t gi\u1eefa hai ta, l\u1eadp ph\u01b0\u01a1ng ch\u1ed7 b\u1ed1n,
\n\u2026 th\u1ebf l\u00e0 ok.<\/p>\n

C\u00d4NG TH\u1ee8C G\u1ea4P \u0110\u00d4I<\/strong><\/p>\n

+Sin g\u1ea5p \u0111\u00f4i = 2 sin cos
\n+Cos g\u1ea5p \u0111\u00f4i = b\u00ecnh cos tr\u1eeb b\u00ecnh sin
\n= tr\u1eeb 1 c\u1ed9ng hai l\u1ea7n b\u00ecnh cos
\n= c\u1ed9ng 1 tr\u1eeb hai l\u1ea7n b\u00ecnh sin
\n+Tang g\u1ea5p \u0111\u00f4i
\nTang \u0111\u00f4i ta l\u1ea5y \u0111\u00f4i tang (2 tang)
\nChia 1 tr\u1eeb l\u1ea1i b\u00ecnh tang, ra li\u1ec1n.<\/p>\n

C\u00e1ch nh\u1edb c\u00f4ng th\u1ee9c: tan(a+b)=(tan+tanb)\/1-tana.tanb l\u00e0<\/strong><\/p>\n

tan m\u1ed9t t\u1ed5ng hai t\u1ea7ng cao r\u1ed9ng
\ntr\u00ean th\u01b0\u1ee3ng t\u1ea7ng tan c\u1ed9ng tan tan
\nd\u01b0\u1edbi h\u1ea1 t\u1ea7ng s\u1ed1 1 ngang t\u00e0ng
\nd\u00e1m tr\u1eeb m\u1ed9t t\u00edch tan tan oai h\u00f9ng
\n
\nC\u00d4NG TH\u1ee8C BI\u1ebeN \u0110\u1ed4I T\u00cdCH TH\u00c0NH T\u1ed4NG<\/strong><\/p>\n

Cos cos n\u1eeda cos-c\u1ed9ng, c\u1ed9ng cos-tr\u1eeb
\nSin sin n\u1eeda cos-tr\u1eeb tr\u1eeb cos-c\u1ed9ng
\nSin cos n\u1eeda sin-c\u1ed9ng c\u1ed9ng sin-tr\u1eeb<\/p>\n

C\u00d4NG TH\u1ee8C BI\u1ebeN \u0110\u1ed4I T\u1ed4NG TH\u00c0NH T\u00cdCH<\/strong><\/p>\n

sin t\u1ed5ng l\u1eadp t\u1ed5ng sin c\u00f4
\nc\u00f4 t\u1ed5ng l\u1eadp hi\u1ec7u \u0111\u00f4i c\u00f4 \u0111\u00f4i ch\u00e0ng
\nc\u00f2n tan t\u1eed c\u1ed9ng \u0111\u00f4i tan (ho\u1eb7c l\u00e0: tan t\u1ed5ng l\u1eadp t\u1ed5ng hai tan)
\nm\u1ed9t tr\u1eeb tan t\u00edch m\u1eabu mang th\u01b0\u01a1ng s\u1ea7u
\ng\u1eb7p hi\u1ec7u ta ch\u1edb lo \u00e2u,
\n\u0111\u1ed5i tr\u1eeb th\u00e0nh c\u1ed9ng ghi s\u00e2u v\u00e0o l\u00f2ng<\/p>\n

M\u1ed9t phi\u00ean b\u1ea3n kh\u00e1c c\u1ee7a c\u00e2u Tan m\u00ecnh c\u1ed9ng v\u1edbi tan ta, b\u1eb1ng sin 2 \u0111\u1ee9a tr\u00ean cos ta cos m\u00ecnh\u2026 l\u00e0<\/strong><\/p>\n

tanx + tany: t\u00ecnh m\u00ecnh c\u1ed9ng l\u1ea1i t\u00ecnh ta, sinh ra hai \u0111\u1ee9a con m\u00ecnh con ta<\/p>\n

tanx \u2013 tan y: t\u00ecnh m\u00ecnh hi\u1ec7u v\u1edbi t\u00ecnh ta sinh ra hi\u1ec7u ch\u00fang, con ta con m\u00ecnh<\/p>\n

C\u00d4NG TH\u1ee8C CHIA \u0110\u00d4I (t\u00ednh theo t=tg(a\/2))<\/strong><\/p>\n

Sin, cos m\u1eabu gi\u1ed1ng nhau ch\u1ea3 kh\u00e1c
\nAi c\u0169ng l\u00e0 m\u1ed9t c\u1ed9ng b\u00ecnh t\u00ea (1+t^2)
\nSin th\u00ec t\u1eed c\u00f3 hai t\u00ea (2t),
\ncos th\u00ec t\u1eed c\u00f3 1 tr\u1eeb b\u00ecnh t\u00ea (1-t^2).<\/p>\n

H\u1ec6 TH\u1ee8C L\u01af\u1ee2NG TRONG TAM GI\u00c1C VU\u00d4NG<\/strong><\/p>\n

Sao \u0110i H\u1ecdc ( Sin = \u0110\u1ed1i \/ Huy\u1ec1n)
\nC\u1ee9 Kh\u00f3c Ho\u00e0i ( Cos = K\u1ec1 \/ Huy\u1ec1n)
\nTh\u00f4i \u0110\u1eebng Kh\u00f3c ( Tan = \u0110\u1ed1i \/ K\u1ec1)
\nC\u00f3 K\u1eb9o \u0110\u00e2y ( Cotan = K\u1ec1\/ \u0110\u1ed1i)<\/p>\n

Sin : \u0111i h\u1ecdc (c\u1ea1nh \u0111\u1ed1i \u2013 c\u1ea1nh huy\u1ec1n)
\nCos: kh\u00f4ng h\u01b0 (c\u1ea1nh \u0111\u1ed1i \u2013 c\u1ea1nh huy\u1ec1n)
\nTang: \u0111o\u00e0n k\u1ebft (c\u1ea1nh \u0111\u1ed1i \u2013 c\u1ea1nh k\u1ec1)
\nCotang: k\u1ebft \u0111o\u00e0n (c\u1ea1nh k\u1ec1 \u2013 c\u1ea1nh \u0111\u1ed1i)<\/p>\n

T\u00ecm sin l\u1ea5y \u0111\u1ed1i chia huy\u1ec1n
\nCosin l\u1ea5y c\u1ea1nh k\u1ec1, huy\u1ec1n chia nhau
\nC\u00f2n tang ta h\u00e3y t\u00ednh sau
\n\u0110\u1ed1i tr\u00ean, k\u1ec1 d\u01b0\u1edbi chia nhau ra li\u1ec1n
\nCotang c\u0169ng d\u1ec5 \u0103n ti\u1ec1n
\nK\u1ec1 tr\u00ean, \u0111\u1ed1i d\u01b0\u1edbi chia li\u1ec1n l\u00e0 ra<\/p>\n

Sin b\u00f9, cos \u0111\u1ed1i, h\u01a1n k\u00e9m pi tang, ph\u1ee5 ch\u00e9o.<\/strong><\/p>\n

+Sin b\u00f9 :Sin(180-a)=sina
\n+Cos \u0111\u1ed1i :Cos(-a)=cosa
\n+H\u01a1n k\u00e9m pi tang :
\nTg(a+180)=tga
\nCotg(a+180)=cotga
\n+Ph\u1ee5 ch\u00e9o l\u00e0 2 g\u00f3c ph\u1ee5 nhau th\u00ec sin g\u00f3c n\u00e0y = cos g\u00f3c kia, tg g\u00f3c n\u00e0y = cotg g\u00f3c kia.<\/p>\n

C\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t h\u01a1n v\u1ec1 vi\u1ec7c h\u01a1n k\u00e9m pi nh\u01b0 sau:<\/strong><\/p>\n

H\u01a1n k\u00e9m b\u1ed9i hai pi sin, cos
\nTang, cotang h\u01a1n k\u00e9m b\u1ed9i pi.
\nSin(a+k.2.180)=sina ; Cos(a+k.2.180)=cosa
\nTg(a+k180)=tga ; Cotg(a+k180)=cotga
\n*sin b\u00ecnh + cos b\u00ecnh = 1
\n*Sin b\u00ecnh = tg b\u00ecnh tr\u00ean tg b\u00ecnh c\u1ed9ng 1.
\n*cos b\u00ecnh = 1 tr\u00ean 1 c\u1ed9ng tg b\u00ecnh.
\n*M\u1ed9t tr\u00ean cos b\u00ecnh = 1 c\u1ed9ng tg b\u00ecnh.
\n*M\u1ed9t tr\u00ean sin b\u00ecnh = 1 c\u1ed9ng cotg b\u00ecnh.
\n(Ch\u00fa \u00fd sin *; cos @ ; tg @ ;cotg * v\u1edbi c\u00e1c d\u1ea5u * v\u00e0 @ l\u00e0 ch\u00fang c\u00f3 li\u00ean quan nhau trong CT tr\u00ean<\/p>\n

DI\u1ec6N T\u00cdCH<\/strong><\/p>\n

Mu\u1ed1n t\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang
\n\u0110\u00e1y l\u1edbn, \u0111\u00e1y b\u00e9 ta mang c\u1ed9ng v\u00e0o
\nR\u1ed3i \u0111em nh\u00e2n v\u1edbi \u0111\u01b0\u1eddng cao
\nChia \u0111\u00f4i k\u1ebft qu\u1ea3 th\u1ebf n\u00e0o c\u0169ng ra.<\/p>\n

Mu\u1ed1n t\u00ecm di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng,
\nC\u1ea1nh nh\u00e2n v\u1edbi c\u1ea1nh ta th\u01b0\u1eddng ch\u1eb3ng sai
\nChu vi ta \u0111\u00e3 h\u1ecdc b\u00e0i,
\nC\u1ea1nh nh\u00e2n v\u1edbi b\u1ed1n c\u00f3 sai bao gi\u1edd.
\nMu\u1ed1n t\u00ecm di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n,
\nPi nh\u00e2n b\u00e1n k\u00ednh, b\u00ecnh ph\u01b0\u01a1ng s\u1ebd th\u00e0nh.<\/p>\n

Nguy\u00ean t\u1eafc \u0111\u1ec3 2 tam gi\u00e1c b\u1eb1ng nhau
\nCon g\u00e0 con, g\u00e2n c\u1ed5 g\u00e1y, c\u00fac c\u00f9 cu
\n(c\u1ea1nh g\u00f3c c\u1ea1nh, g\u00f3c c\u1ea1nh g\u00f3c, c\u1ea1nh c\u1ea1nh c\u1ea1nh)<\/p>\n

Nh\u1eefng c\u00f4ng th\u1ee9c l\u01b0\u1ee3ng gi\u00e1c kh\u00f3 nh\u1eb1n \u0111\u00e3 \u0111\u01b0\u1ee3c \u0111\u01a1n gi\u1ea3n h\u00f3a b\u1eb1ng nh\u1eefng c\u00e2u th\u01a1 quen thu\u1ed9c, g\u1ea7n g\u0169i, gi\u00fap h\u1ecdc sinh d\u1ec5 d\u00e0ng v\u01b0\u1ee3t qua d\u1ea1ng b\u00e0i l\u01b0\u1ee3ng gi\u00e1c.<\/p>\n","protected":false},"excerpt":{"rendered":"

L\u01b0\u1ee3ng gi\u00e1c l\u00e0 d\u1ea1ng b\u00e0i xu\u1ea5t hi\u1ec7n trong \u0111\u1ec1 thi m\u00f4n to\u00e1n THPT Qu\u1ed1c gia nh\u01b0ng thay v\u00ec nh\u1edb nh\u1eefng c\u00f4ng th\u1ee9c kh\u00f4 khan n\u00e0y, t\u1ea1i sao b\u1ea1n kh\u00f4ng th\u1eed bi\u1ebfn n\u00f3 th\u00e0nh nh\u1eefng c\u00e2u th\u01a1 nh\u1ec9. H\u00c0M S\u1ed0 L\u01af\u1ee2NG GI\u00c1C B\u1eaft \u0111\u01b0\u1ee3c qu\u1ea3 tang Sin n\u1eb1m tr\u00ean cos (tan@ = sin@:cos@) Cotang d\u1ea1i d\u1ed9t […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[],"yoast_head":"\n