Trigonometric Identities
sin(A+ B) = sinAcosB+ cosAsinB (6) sin(A B) = sinAcosB cosAsinB (7) tan(A+ B) = tanA+ tanB 1 tanAtanB (8) tan(A B) = tanA tanB 1 + tanAtanB (9) cos2 = cos2 sin2 = 2cos2 1 = 1 2sin2 (10) sin2 = 2sin cos (11) tan2 = 2tan 1 tan2 (12) Note that you can get (5) from (4) by replacing B with B, and using the fact that cos( B) = cosB(cos is even) and
Trigonometric Identities - Miami
[sin(x+ y) + sin(x y)] Sum-to-Product Formulas sinx+ siny= 2sin x+y 2 cos x y 2 sinx siny= 2sin x y 2 cos x+y 2 cosx+ cosy= 2cos x+y 2 cos x y 2 cosx cosy= 2sin x+y 2 sin x y 2 The Law of Sines 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
44 Trigonometrical Identities
sinA cosA secA = 1 cosA cosecA = 1 sinA cotA = cosA sinA = 1 tanA sin(A±B)=sinAcosB ±cosAsinB cos(A±B)=cosAcosB ∓sinAsinB tan(A±B)= tanA±tanB 1∓tanAtanB 2sinAcosB =sin(A+B)+sin(A−B) 2cosAcosB =cos(A−B)+cos(A+B) 2sinAsinB =cos(A−B)−cos(A+B) sin 2A+cosA =1 1+cot 2A =cosecA, tanA+1=sec2 A cos2A =cos2 A−sin2 A =2cos 2A−1=1
Formulas from Trigonometry
Formulas from Trigonometry: sin 2A+cos A= 1 sin(A B) = sinAcosB cosAsinB cos(A B) = cosAcosB tansinAsinB tan(A B) = A tanB 1 tanAtanB sin2A= 2sinAcosA cos2A= cos2 A sin2 A tan2A= 2tanA
Lesson 183: Triangle Trigonometry
b sin sin b sin sin b sinC c sinB b ≈ ° ⋅ ° = ° = ° = Thus b is the longest side of this triangle Since ∠B is the largest angle in the triangle, this result is consistent Triangle 2: If ∠A =123° = ° ∠ = − − 32 B 180 123 25 To find b: 8 5 25 6 8 32 25 6 8 32 b sin sin b sin sin b sinC c sinB b
Complex numbers and Trigonometric Identities
• Distributing exponents was his only sin • But that’s enough to do an algebra student in • An example, his demise should serve, • For other students who haven’t heard, • Distributing exponents is a sin • It’s enough to do an algebra student in • • Donald E Brook • Mt San Antonio College • ????????+ ????????
Trignometrical Formulae Standard Integrals
2sinA cosB = sin(A+B)+sin(A−B) 2cosA sinB = sin(A+B)−sin(A−B) 2cosA cosB = cos(A+B)+cos(A−B) 2sinA sinB = cos(A−B)−cos(A+B) Hyperbolic Functions sinhx = ex −e−x 2, coshx = ex +e−x 2 Standard Derivatives f(x) f0(x) x nnx −1 sinax acosax cosax −asinax tanax asec2 ax e axae lnx 1 x sinhax acoshax coshax asinhax uv u0 v +uv0 u
Formulas - Hong Kong University of Science and Technology
(a+b)sin 1 2 (b−a) sinacosb= 1 2 (sin(a−b)+sin(a+b)) sinasinb= 1 2 (cos(a−b)−cos(a+b)) cosacosb= 1 2 (cos(a−b)+cos(a+b)) sin(a±b) = sinacosb±cosasinb cos(a±b) = cosacosb∓sinasinb Fourier series Fourier series of f(x) defined on [−L,−L]: 1 2 a0 + X∞ n=1 (an cos(nπx/L)+bn sin(nπ/L)) where an = 1 L Z L −L f(x)cos(nπx/L
Trigonometric Identities Revision : 1
Finally, from equations (2) and (3) we can obtain an identity for tan(A+B): tan(A+B) = sin(A+B) cos(A+B) = sinAcosB +cosAsinB cosAcosB −sinAsinB Now divide numerator and denominator by cosAcosB to obtain the identity we wanted: tan(A+B) = tanA+tanB 1−tanAtanB (16) We can get the identity for tan(A − B) by replacing B in (16) by −B and
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