sinA a = sinB b = sinC c Suppose you are given two sides, a;band the angle Aopposite the side A The height of the triangle is h= bsinA Then 1 If ahand a
Trignometrical Formulae sin(A+B) = sinA cosB +cosA sinB sin(A−B) = sinA cosB −cosA sinB cos(A+B) = cosA cosB −sinA sinB cos(A−B) = cosA cosB +sinA sinB
sina±sinb= 2sin 1 2 (a±b)cos 1 2 (a∓b) cosa+cosb= 2cos 1 2 (a+b)cos 1 2 (a−b) cosa−cosb= 2sin 1 2 (a+b)sin 1 2 bn sin(nπx/L) where bn = 2 L Z L 0 f(x)sin
Dividing both sides by sinAsinB results in: sinB b sinA a = Now drop a perpendicular line BE of length k from the vertex B to the side AC (Diagram 2 ) By right triangle trigonometry: k csinA c k sinA = = and k asinC a k sinC = = Hence: sinC c sinA a a sinC c sinA = = Since: sinB b sinA a = and sinC c sinA a = then: sinC c sinB b sinA a = = The
cos(A B) = cosAcosB tansinAsinB tan(A B) = A tanB 1 tanAtanB sin2A= 2sinAcosA cos2A= cos2 A sin2 A tan2A= 2tanA 1 2tan A sin A 2 = q 1 cosA 2 cos A 2 = q 1+cos A 2 tan 2 = sinA 1+cosA sin2 A= 1 2 21 2 cos2A cos A= 1 2 + 1 2 cos2A sinA+sinB= 2sin 1 2 (A+B)cos 1 2 (A 1B) sinA sinB= 2cos 1 2 (A+B)sin 2 (A B) cosA+cosB= 2cos 1 2 (A+B)cos 1 2 (A B
A= B= Sum and product formulae cosA+ cosB= 2cos A+ B 2 cos A B 2 (13) cosA cosB= 2sin A+ B 2 sin A B 2 (14) sinA+ sinB= 2sin A+ B 2 cos A B 2 (15) sinA sinB= 2cos A+ B 2 sin A B 2 (16) Note that (13) and (14) come from (4) and (5) (to get (13), use (4) to expand cosA= cos(A+ B 2 + 2) and (5) to expand cosB= cos(A+B 2 2), and add the results)
sina-sinb=2cos 2 a b sin 2 a b cosa+cosb = 2cos cos cosa-cosb = -2sin sin tana+tanb= a b a b s s ( ) 积化和差 sinasinb = - 2 1 [cos(a+b)-cos(a-b)] cosacosb =
If we let A = B in equations (2) and (3) we get the two identities sin2A = 2sinAcosA, (12) cos2A = cos2 A−sin2 A (13) 2 6 Identities for sine squared and cosine
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cos( ) cos cos sin sina b a b a b+ = − cos( ) cos cos sin sina b a b a b− = + sin( ) sin cos cos sina b a b a b+ = + sin( ) sin cos cos sina b a b a b− = −Taille du fichier : 150KB
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正弦定理 a/sinA=b/sinB=c/sinC=2R 注： 其中 R 表示三角形的外接圆半径 余弦定理 b2=a2+c2-2accosB 注：角B 是边a 和边c 的夹角 正切定理: [(a+b)/(a-b)]={[Tan(a+b)/2]/[Tan(a-b)/2]} 圆的标准方程 (x-a)2+(y-b)2=r2 注：（a,b）是圆心坐标 圆的一般方程 x2+y2+Dx+Ey+F=0 注：D2+E2-4F>0
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cos(A + B) = cos A cos B ? sin A sin B that cos(?B) = cos B (cos is even) and sin(?B) = ? sin B (sin is odd). Similarly (7) ... A = B = ?.
Trignometrical Formulae sin(A + B) = sinA cosB + cosA sinB sin(A ? B) = sinA cosB ? cosA sinB cos(A + B) = cosA cosB ? sinA sinB.
sin (A+B)=sinA cos B+ cos Asin B sin (A-B) - sin A cosB -cos AsinB. 636 Trigonometry - Equations and Identities Lesson #6: Sum and Difference Identities.
sin C. Example. In triangle ABC B = 21?
sin acos b =}[sin (a+ b )+ sin (a - b )] sin a sin b =}[oos(a-b)-cos(a+b)] ... TRIGONOMETRIC VALUES FOR COMMON ANGLES sin 8.
a3 + b3 = (a + b)(a2 - ab + b2). sin t sect = 1 cos t cott = 1 tan t. = cos t sin t . Table 6.1: Reciprocal and Quotient Identities.
cos(?1 + ?2) = cos ?1 cos ?2 ? sin ?1 sin ?2 sin(?1 + ?2) = sin ?1 cos where a b are real numbers
Cuboid : Total surface area = 2 (ab + bh + lh); Volume = lbh. sin ? ? . sin2? + cos2? = 1;? sin2? = 1- cos2?; cos2? = 1- sin2?;.
Suppose both AB and AC have a length of ?. 2 radians. The Spherical Law of Cosines says cos(BC) = cos(AB)cos(AC) + sin(AB)sin(AC)cos(ZBAC).
sin(A $ B) ? sinAcosB $ cosAsinB sin .-/. ( cos ? $ '. ( ? %sin? sin ? $ '. ( ? $cos? sin (90o ?) ? cos? cos (90o ... [cos (A B) + cos (A + B)].
Trignometrical Formulae sin(A + B) = sinA cosB + cosA sinB sin(A ? B) = sinA cosB ? cosA sinB cos(A + B) = cosA cosB ? sinA sinB cos(A ? B) = cosA cosB
cos(A + B) = cos A cos B ? sin A sin B that cos(?B) = cos B (cos is even) and sin(?B) = ? sin B (sin is odd) Similarly (7) A = B = ?
sin(x) cos(x) définie si x = ? 2 (?) cotan(x) = 1 tan(x) = cos(x) sin(x) définie si x =0 (?) cos2(x) + sin2(x) = 1 1 + tan2(x) = 1 cos2(x) si x =
cos (a + b) = cos(-b) cos a + sin(-b) sina or cos (-b) = cos b et sin(-b) = - sin b On obtient donc : cos (a + b) = cos b cos a – sinb sina • sin(ab)
S'il existe ? tel que cos(?) = a et sin(?) = b le théorème 3 montre que a2 +b2 = 1 Réciproquement si a2 + b2 = 1 le point M(a b) est un point du cercle
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Le but de cette note est de simplifier la discussion des for* ' mules qui donnent sin (a±b) cos {a±b) On n'a besoin d'y considérer aucune relation de
On aurait aussi dès lors : DF==ADcosD mais DF = AD sin A ; donc sin A = cos D = cos (90°— A) ; ce qui montre que le sinus d'un angle est égal au cosinus du
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Trigonometric Identities