This article explains in detail how De Moivre's Theorem may be used to find 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 means 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.
De Moivre’s Theorem For cos3x and sin3x
(cosx + i.sinx)^3 = cos3x + i.sin3x,
where i = √(-1)
The Binomial Theorem may be used to expand the left hand side:
(cosx + i.sinx)³ = cos³x + 3. cos²x.i.sinx + 3.cosx.(i.sinx)² + (i.sinx)³
(cosx + i.sinx)³ = cos³x + 3.i.cos²x.sinx – 3.cosx.sin²x – i.sin³x (Equation 1)
Noting that if two complex numbers are identical, then the real parts of those numbers must be the same, and the complex ("imaginary") parts of those numbers are the same. i.e.
if
a + i.b = c + i.d, then
a = c, and b = d.
With this in mind, it must follow that
cos3x = real part of
(cos³x + 3.i.cos²x.sinx – 3.cosx.sin²x – i.sin³x)
So, cos3x = cos³x– 3.cosx.sin²x
It is customary to state cos3x, 4x etc as a function of cosx. Using the common trigonometric identity
sin²x + cos²x = 1, and restating it as
sin²x = 1 - cos²x, then cos3x becomes
cos³x– 3.cosx.sin²x
= cos³x – 3. cosx (1 – cos²x)
= cos³x – 3.cosx + 3.cos³x
= 4.cos³x – 3.cosx
As an example, take the case of x = π / 4 (or 45°):
cos(π / 4) = 0.7071, so cos(3π / 4)
= 4.(0.7071)³ - 3.(0.7071)
= 4×0.3536 – 3×0.7071
= -0.7071
This is equal to cos(3π / 4) as expected.
To find the formula for sin3x, Equation 1 is used:
(cosx + i.sinx)^3 = cos³x + 3.i.cos²x.sinx – 3.cosx.sin²x – i.sin³x (Equation 1)
sin3x = imaginary part of (cos³x + 3.i.cos²x.sinx – 3.cosx.sin²x – i.sin³x)
= 3.cos²x.sinx – sin³x
It is often common to express this in terms of sinx, so
sin3x = 3.(1 – sin²x). sinx – sin³x
= 3.sinx – 3.sin³x – sin³x
= 3.sinx – 4.sin³x
Verifying this with x = π / 6 (or 30°)
sin(π/6) = 0.5, and sin³(π/6) = 0.125,
so sin(3×π/6) = sin(π/2)
= 3×0.5 – 4×0.125
= 1.5 – 0.5
= 1, which is equal to sin(π/2) as expected.
De Moivre’s Theorem For cos4x and sin4x
The following steps are used, without further explanation, as these have been described already for cos3x and sin3x:
State De Moivre’s Theorem
Expand the power term using the Binomial Theorem to get the overall expression
Gather real and imaginary terms
Set cos4x equal to the real part of the expression
Set sin4x equal to the imaginary part of the expression
Eliminate sinx terms in the expression for cos4x where possible
Eliminate cosx terms in the expression for sin4x where possible
(cosx + i.sinx)^4 = cos4x + i.sin4x,
where i = √(-1)
(cosx + i.sinx)^4 =
(cosx)^4 + 4.cos³x.i.sinx + 6.cos²x.(i.sinx)² + 4.cosx.(i.sinx)³ + (i.sinx)^4
= (cosx)^4 + 4.cos³x.i.sinx + 6.cos²x.(-1).sin²x + 4.cosx.i.(-1).sin³x + (sinx)^4 (Equation 2)
So cos4x = (cosx)^4 –6. cos²x.sin²x + (sinx)^4
= (cosx)^4 –6. cos²x.(1 - cos²x) + (1 - cos²x)²
= (cosx)^4 – 6.cos²x + 6.(cosx)^4 + 1 –2.cos²x + (cosx)^4
= 8. (cosx)^4 – 8.cos²x +1
From Equation 2,
Sin4x = imaginary part of (cosx)^4 + 4.cos³x.i.sinx + 6.cos²x.(-1).sin²x + 4.cosx.i.(-1).sin³x + (sinx)^4
sin4x = 4.cos³x.sinx + 4.cosx.(-1).sin³x
= 4.cos³x.sinx – 4.cosx.sin³x
= 4.cosx.sinx.((1 – sin²x) – sin²x)
= 4.cosx.sinx.(1 – 2.sin²x)
Cos3x, cos4x, sin3x and sin4x Summary
The formulas for the four stated functions have been derived using De Moivre's Theorem. Using the steps shown in this article, it is possible to easily find the formula for the sine or cosine of 5x, 6x etc. A further tip is to note that where only one of the sine / cosine formulas is needed, there is no need to evaluate real / imaginary parts of the binomial expansion. Fig 1 and Fig 2 show how to find cos5x and sin5x respectively.
Cos3x, cos4x, sin3x and sin4x References
Most of the work here is derived from the articles "De Moivres Theorem Description With Examples and Application" and "Trigonometric Identities Lesson and Manipulating Trig Functions".
Martin Bell - Martin holds a B.Sc. degree in chemical engineering, and an M.Sc. degree in electronics and computing. He has spent more than 25 years ...
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