Cos of multiples of x, in terms of cos x

As you know from Trig Equiv and Sin or Cos 3x,

cos 2x = -sin²+cos², 
cos 3x = cos³x - 3sin²x cos x

If you keep going, you'll see the cosines of all the multiples of x have only even powers of sin x, so you can always use
sin²x=(1-cos²x) to express cos nx in terms of cos x:

cos 2x = -1 + 2cos²x, 
cos 3x = -3cos x + 4cos³x

As the multiples of x get higher, it gets harder and harder to multiply out all the factors of (1-cos²x), so I looked for a way to express cos nx in terms of cos x, cos (n-1)x, and cos (n-2)x.

Let y = (n-1)x.  Then nx is y+x and (n-2)x is y-x.

You know that

cos y+x = cos y cos x - sin y sin x, and
cos y-x = cos y cos x + sin y sin x

So the sum  of cos y+x and cos y-x is 2 cos y cos x.  So we get this useful trig equivalence involving just cosines:

cos y+x = 2 cos y cos x - cos y-x

Substituting (n-1)x in place of y, nx in place of y+x, and (n-2)x in place of y-x, we get this:

cos nx = 2 cos[(n-1)x] cos x - cos[(n-2)x]

Now that we know cos 2x and cos 3x, let's see how to use this fact to find cos 4x:

cos 2x = -1 + 2cos²x, 
cos 3x = -3cos x + 4cos³x
cos 4x = 2 cos 3x cos x - cos 2x
cos 4x = 2 (-3cos x + 4cos³x) cos x - (-1 + 2cos²x)
cos 4x = (-6cos²x + 8cos⁴x) - ( -1 + 2cos²x)
cos 4x = 1 - 8cos² x + 8cos⁴x

I'm far too lazy to keep doing these by hand, so I wrote a spreadsheet that calculates cos of all the multiples of x:

cos 1x = cos 
cos 2x = -1+2cos²  
cos 3x = -3cos+4cos³  
cos 4x = 1-8cos²+8cos⁴  
cos 5x = 5cos-20cos³+16cos⁵  
cos 6x = -1+18cos²-48cos⁴+32cos⁶  
cos 7x = -7cos+56cos³-112cos⁵+64cos⁷  
cos 8x = 1-32cos²+160cos⁴-256cos⁶+128cos⁸  
cos 9x = 9cos-120cos³+432cos⁵-576cos⁷+256cos⁹  
cos 10x = -1+50cos²-400cos⁴+1120cos⁶-1280cos⁸+512cos¹⁰  
cos 11x = -11cos+220cos³-1232cos⁵+2816cos⁷-2816cos⁹+1024cos¹¹  
cos 12x = 1-72cos²+840cos⁴-3584cos⁶+6912cos⁸-6144cos¹⁰+2048cos¹²  
cos 13x = 13cos-364cos³+2912cos⁵-9984cos⁷+16640cos⁹-13312cos¹¹+4096cos¹³  
cos 14x = -1+98cos²-1568cos⁴+9408cos⁶-26880cos⁸+39424cos¹⁰-28672cos¹²+8192cos¹⁴  
cos 15x = -15cos+560cos³-6048cos⁵+28800cos⁷-70400cos⁹+92160cos¹¹-61440cos¹³+16384cos¹⁵  
cos 16x = 1-128cos²+2688cos⁴-21504cos⁶+84480cos⁸-180224cos¹⁰+212992cos¹²-131072cos¹⁴+32768cos¹⁶  
cos 17x = 17cos-816cos³+11424cos⁵-71808cos⁷+239360cos⁹-452608cos¹¹+487424cos¹³-278528cos¹⁵+65536cos¹⁷  
cos 18x = -1+162cos²-4320cos⁴+44352cos⁶-228096cos⁸+658944cos¹⁰-1118208cos¹²+1105920cos¹⁴-589824cos¹⁶+131072cos¹⁸  

How was this table created?

A good question.  I'm glad you asked!  The coefficients of these formulas form a kind of "Pascal's Triangle", except we use a slightly different calculation from that of Pascal's Triangle, which gives the binomial coefficients.

Each coefficient is the twice the one above and to the left of it minus the one two rows above it.  That formula comes directly from this equivalence, which we developed, above:

cos nx = 2 cos[(n-1)x] cos x - cos[(n-2)x]

I'll give you the first few rows of the triangle:

cos nx

1
    1
-1      2
    -3       4
1      -8        8
     5      -20        16
-1      18      -48        32
     -7      56       -112        64
1      -32      160       -256          128
      9     -120       432       -576          256
-1      50      -400       1120        -1280         512
    -11     220       -1232      2816        -2816         1024
1      -72       840       -3584         6912        -6144         2048
     13    -364        2912      -9984         16640       -13312       4096
-1      98      -1568       9408       -26880       39424        -28672         8192
    -15     560       -6048       28800       -70400       92160       -61440        16384
1      -128      2688      -21504       84480      -180224       212992       -131072       32768
     17      -816      11424      -71808      239360      -452608      487424      -278528        65536
-1      162      -4320      44352      -228096      658944      -1118208      1105920      -589824      131072

Next, a similar table for sines.

Related Pages in this website

Special Angles

Special Angles, part 2 uses the information presented, above, to solve a puzzle: find the sum of sec 40�, sec 80�, and sec 160�.

Trig Equivalences

Sin or Cos 3x, 4x, etc. -- trig functions of any multiple of an angle.

Sin of multiples of x, in terms of cos x

d/dx (sin x) = cos x, in the calculus section of this website

Hyperbolic Functions -- sinh(x) and cosh(x), which, together with exp(x) and the circular functions sin(x) and cos(x) form a family of functions.

The webmaster and author of this Math Help site is Graeme McRae.