Trigonometry Cosine, Sine and Tangent of Multiple Angles (Chebyshev's
Whilst De Moivre's Theorem for Multiple Angles enables us to compute a
sine or cosine of a multiple angle directly, for the cosine we need to
convert powers of sine to cosines (and similarly for the sine).
However, Chebyshev's Method gives the formula in the required form for
the cosine, and, for sines, requires the conversion of cosine squared
to sines only (as opposed to converting higher powers using DeMoivre). However, there is an algorithm which does this more
directly. Computing the tangents can become complicated, as n
increases, so the method for tangents is particularly valueable. I
to the method as Chebyshev's Method, because Pafnuty
Chebyshev (Tchebysheff) (1821-1894), a Russian Mathematician,
certainly knew this formula (and for any later reference, it is better to give it some name!). This is a recursive method. See also:
We compute the cosine for nx from the cosines of (n-1) and (n-2) as follows:
Similarly we compute the sine of nx from the sines of (n-1)x and (n-2)x
For the tangent, we have: [1.3] where H/K=tan (n-1)x
Starting from cos(1·x)=cos x, and cos2x=2cos2x-1, using formula 1.1, we can compute cos 3x=2cos x(2cos2x−1)−cos x =4·cos3x−2·cos x−cos x =4·cos3x−3·cos x ■
Starting from sin(1·x)=sin x, and sin2x=2·sin x·cos x, using formula 1.2, we can compute sin 3x=2·cos x·(2·sin x·cos x)−sin x =4·cos2x·sin x−sin x =4·(1−sin2x)·sin x−sin x =3·sin x−4sin3x ■
we wish to find tan2x; that is the multiple angle formula for the
tangent when n=2, and we know tan (n-1)x=tan x. Therefore n=2,
H=tan x, and K=1, using formula 1.3, reproduced below: [1.3] And substituting our values, we find: tan 2x=(tan x+tan x·1)/(1-tan x·tan x) =2·tan x/(1-tan2x) ■ If we continue, with n=3, H=2·tan x, K=1−tan2x, using the formula we have: tan 3x=(2·tan x+tan x·(1−tan2x))/(1−tan2x−(2·tan x)tan x)
=(3·tan x−tan3x)/(1−3·tan2x) ■ This formula uses the previous term only (as apposed to the previous two terms for the sine and cosine formulae).
That we have chosen to claim to prove
the formula for all integers, n, neither implies nor does not imply the
truth of the equations for rational, real or complex numbers.
We wish to prove that: [1.1, repeated] for all integers, n.
To do this we use the compound angle formula to show that the right-hand side of Equation 1.1 is actually cos(nx).
Expanding the right-hand side, using the compound angle formula for sine and cosine: cos nx=[2·cos(nx)cos (x) + 2·sin(nx)sin x]·cos (x)−[cos(nx)cos2x+sin(nx)sin2x]
Writing cos2x as 2·cos2x−1, and sin2x as 2·sin x·cosx, and multiplying out some brackets: =2·cos(nx)cos2(x)+ 2·sin(nx)sin x·cos (x)−cos(nx)·(2·cos2x−1)−sin(nx)·2·sinx·cosx
Multiplying out further brackets =2·cos(nx)cos2(x)+ 2·sin(nx)sin x·cos (x)−2·cos(nx)·cos2x+cos(nx)−2·sin(nx)·sinx·cosx
Grouping similar terms, most of which cancel: =cos(nx)
We wish to prove that: [1.2, repeated] for all integers n.
Expanding the right-hand side of Equation 1.2, using the compound angle formula for sine and cosine: sin nx=[2·sin(nx)cos (x) + 2·cos(nx)sin x]·cos (x)−[sin(nx)cos2x+cos(nx)sin2x]
Writing cos2x as 1−2·sin2x, and sin2x as 2·sin x·cosx =[2·sin(nx)cos (x) + 2·cos(nx)sin x]·cos (x)−sin(nx)(1−2·sin2x)−2·sin x·cos x·cos(nx) =2·sin(nx)cos2 (x)+ 2·cos(nx)sin x·cos (x)−sin(nx)+2·sin2x·sin(nx)−2·sin x·cos x·cos(nx)