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| He hears the commendation, not of himself, but more sweet, of that character he seeks, in every word that is said concerning character, yea, further, in every fact and circumstance, in the running river and the rustling corn. Praise is looked, homage tendered, love flows from mute nature, from the mountains and the lights of the firmament.
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| Wysłany: Pon 7:18, 30 Maj 2011 Temat postu: Retro Jordans De Moivres Theorem Examples cos3x, s |
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| (cosx + i.sinx)³ = cos³x + – – i.sin³x (Equation 1)
So, cos3x = cos³x– It is customary to state cos3x, 4x etc as a function of cosx. Using the common trigonometric identity As one sample, take the circumstance of x = / 4 (or 45): (cosx)^4 + + + + (i.si = 1, which is equal to sin(π/2) as expected. cos3x = real part of (cosx + i.sinx)^4 = (cos³x + – – i.sin³x) (cosx + i.sinx)³ = cos³x + 3. + + (i.sinx)³ a + i.b = c + i.d, then De Moivre’s Theorem For cos4x and sin4x De Moivre’s Theorem For cos3x and sin3x cos(π / 4) = 0.7071, so cos(3π / 4) It is often general to express this in terms of sinx, so The following treads are used Retro Jordans, without beyond explanation, for these have been narrated yet for cos3x and sin3x: Eliminate cosx terms in the statement as sin4x where possible (cosx + i.sinx)^4 = cos4x + i.sin4x, sin3x = 3.(1 – sin²x). sinx – sin³x a = c, and b = d. Expand the power term using the Binomial Theorem to obtain the overall expression Set sin4x equal to the imaginary part of the expression Eliminate sinx terms in the expression for cos4x where possible The Binomial Theorem may be used to distend the left hand side: To find the formula for sin3x, Equation 1 is used: sin(π/6) = 0.5, and sin³(π/6) = 0.125, Gather real and imaginary terms so sin(3×π/6) = sin(π/2) Set cos4x equal to the real part of the expression This is equal to cos(3 / 4) as expected. where i = √(-1) = – sin³x = cos³x – 3.cosx + 3.cos³x Noting that if two complex mathematics are alike, then the real parts of those numbers must be the same, and the complex ("fantastic") parts of those numbers are the same. Read on Trigonometric Identity Advanced Example Trigonometric Identities Lesson and Manipulating Trig Functions Trigonometry Sin(a+b) Cos(a+b) Sin(a)+Sin(b) Cos(a)+Cos(b) if With this in mind, it must follow that Verifying this with x = / 6 (alternatively 30) sin²x + cos²x = 1, and restating it as cos³x– = cos³x – 3. cosx (1 – cos²x) where i = √(-1) = 3.sinx – 4.sin³x (cosx + i.sinx)^3 = cos³x + – – i.sin³x (Equation 1) State De Moivre’s Theorem (cosx + i.sinx)^3 = cos3x + i.sin3x, = 3×0.5 – 4×0.125 = 1.5 – 0.5 sin3x = imaginary part of (cos³x + – – i.sin³x) = -0.7071 sin²x = 1 - cos²x, then cos3x becomes This article explains in detail how De Moivre's Theorem may be accustom to detect the cosine and sine of 3x and 4x. Formulas for cos2x and sin2x are derived in the article "De Moivres Theorem Description With Examples and Application". Note that cos²x method the square of the cosine of x, (cosx)². Fig 1 and Fig 2 show how to find cos5x and sin5x using De Moivre's Theorem respectively. = 3.sinx – 3.sin³x – sin³x = 4.cos³x – 3.cosx = - = 4×0.3536 – 3×0.7071 |
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