Sin or Cos of larger multiples of x

As you know from Trig Equiv,

cos x+y = cos x cos y - sin x sin y
sin x+y = sin x cos y + cos x sin y
tan x+y = (tan x + tan y) / (1 - tan x tan y)

So the double angle formulas follow:

cos 2x = cos²x - sin²x = 1 - 2sin²x = 2cos²x - 1
sin 2x = 2 sin x cos x
tan 2x = 2tan x / (1 - tan² x)

Larger multiples are obtained the same way:

cos 3x = cos 2x+x = cos 2x cos x - sin 2x sin x
   = (cos²x - sin²x) cos x - (2 sin x cos x) sin x
   = cos³x - sin²x cos x - 2 sin²x cos x
   = cos³x - 3sin²x cos x

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

cos 1x = cos
cos 2x = -sin²+cos²
cos 3x = -3cos sin²+cos³
cos 4x = sin⁴-6cos²sin²+cos⁴
cos 5x = 5cos sin⁴-10cos³sin²+cos⁵
cos 6x = -sin⁶+15cos²sin⁴-15cos⁴sin²+cos⁶
cos 7x = -7cos sin⁶+35cos³sin⁴-21cos⁵sin²+cos⁷
cos 8x = sin⁸-28cos²sin⁶+70cos⁴sin⁴-28cos⁶sin²+cos⁸
cos 9x = 9cos sin⁸-84cos³sin⁶+126cos⁵sin⁴-36cos⁷sin²+cos⁹
cos 10x = -sin¹⁰+45cos²sin⁸-210cos⁴sin⁶+210cos⁶sin⁴-45cos⁸sin²+cos¹⁰
cos 11x = -11cos sin¹⁰+165cos³sin⁸-462cos⁵sin⁶+330cos⁷sin⁴-55cos⁹sin²+cos¹¹
cos 12x = sin¹²-66cos²sin¹⁰+495cos⁴sin⁸-924cos⁶sin⁶+495cos⁸sin⁴-66cos¹⁰sin²+cos¹²
cos 13x = 13cos sin¹²-286cos³sin¹⁰+1287cos⁵sin⁸-1716cos⁷sin⁶+715cos⁹sin⁴-78cos¹¹sin²+cos¹³
cos 14x = -sin¹⁴+91cos²sin¹²-1001cos⁴sin¹⁰+3003cos⁶sin⁸-3003cos⁸sin⁶+1001cos¹⁰sin⁴-91cos¹²sin²+cos¹⁴
cos 15x = -15cos sin¹⁴+455cos³sin¹²-3003cos⁵sin¹⁰+6435cos⁷sin⁸-5005cos⁹sin⁶+1365cos¹¹sin⁴-105cos¹³sin²+cos¹⁵
cos 16x = sin¹⁶-120cos²sin¹⁴+1820cos⁴sin¹²-8008cos⁶sin¹⁰+12870cos⁸sin⁸-8008cos¹⁰sin⁶+1820cos¹²sin⁴-120cos¹⁴sin²+cos¹⁶
cos 17x = 17cos sin¹⁶-680cos³sin¹⁴+6188cos⁵sin¹²-19448cos⁷sin¹⁰+24310cos⁹sin⁸-12376cos¹¹sin⁶+2380cos¹³sin⁴-136cos¹⁵sin²+cos¹⁷
cos 18x = -sin¹⁸+153cos²sin¹⁶-3060cos⁴sin¹⁴+18564cos⁶sin¹²-43758cos⁸sin¹⁰+43758cos¹⁰sin⁸-18564cos¹²sin⁶+3060cos¹⁴sin⁴-153cos¹⁶sin²+cos¹⁸

sin 1x = sin
sin 2x = 2cos sin
sin 3x = -sin³+3cos²sin
sin 4x = -4cos sin³+4cos³sin
sin 5x = sin⁵-10cos²sin³+5cos⁴sin
sin 6x = 6cos sin⁵-20cos³sin³+6cos⁵sin
sin 7x = -sin⁷+21cos²sin⁵-35cos⁴sin³+7cos⁶sin
sin 8x = -8cos sin⁷+56cos³sin⁵-56cos⁵sin³+8cos⁷sin
sin 9x = sin⁹-36cos²sin⁷+126cos⁴sin⁵-84cos⁶sin³+9cos⁸sin
sin 10x = 10cos sin⁹-120cos³sin⁷+252cos⁵sin⁵-120cos⁷sin³+10cos⁹sin
sin 11x = -sin¹¹+55cos²sin⁹-330cos⁴sin⁷+462cos⁶sin⁵-165cos⁸sin³+11cos¹⁰sin
sin 12x = -12cos sin¹¹+220cos³sin⁹-792cos⁵sin⁷+792cos⁷sin⁵-220cos⁹sin³+12cos¹¹sin
sin 13x = sin¹³-78cos²sin¹¹+715cos⁴sin⁹-1716cos⁶sin⁷+1287cos⁸sin⁵-286cos¹⁰sin³+13cos¹²sin
sin 14x = 14cos sin¹³-364cos³sin¹¹+2002cos⁵sin⁹-3432cos⁷sin⁷+2002cos⁹sin⁵-364cos¹¹sin³+14cos¹³sin
sin 15x = -sin¹⁵+105cos²sin¹³-1365cos⁴sin¹¹+5005cos⁶sin⁹-6435cos⁸sin⁷+3003cos¹⁰sin⁵-455cos¹²sin³+15cos¹⁴sin
sin 16x = -16cos sin¹⁵+560cos³sin¹³-4368cos⁵sin¹¹+11440cos⁷sin⁹-11440cos⁹sin⁷+4368cos¹¹sin⁵-560cos¹³sin³+16cos¹⁵sin
sin 17x = sin¹⁷-136cos²sin¹⁵+2380cos⁴sin¹³-12376cos⁶sin¹¹+24310cos⁸sin⁹-19448cos¹⁰sin⁷+6188cos¹²sin⁵-680cos¹⁴sin³+17cos¹⁶sin
sin 18x = 18cos sin¹⁷-816cos³sin¹⁵+8568cos⁵sin¹³-31824cos⁷sin¹¹+48620cos⁹sin⁹-31824cos¹¹sin⁷+8568cos¹³sin⁵-816cos¹⁵sin³+18cos¹⁷sin

You may notice that the cos table has all even sine powers, so it would be kind of nice to reformat it using sin²x=1-cos²x to eliminate all the sines.  You could do that, but it turns out there's an easier way.  See Cos of multiples of x, in terms of cos x 

Also, you might notice that the sin table is such that a sin can be factored out of every right-hand-side, leaving all even sine powers, so all but one sine can be eliminated from every right-hand side.  See Sin of multiples of x, in terms of cos x

How was this table created?

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

First, notice that the coefficients of these formulas are arranged in increasing order of cosine-power, with zeros omitted.  I'll put back the zeros, and give you the first few rows of the two triangles:

cos nx

1
0   1
-1   0   1
0   -3   0   1
1   0   -6   0   1
0   5   0   -10   0   1
-1   0   15   0   -15   0   1

sin nx

1
1   0
0   2   0
-1   0   3   0
0   -4   0   4   0
1   0   -10   0   5   0
0   6   0   -20   0   6   0

Just as in Pascal's triangle, these triangles form each element from a combination of two elements above it.  However, one of the two elements is from the other table!  Here's how it works.  Each element of the cosine triangle is the element above and to the left minus the element that is above and to the right, except from the other table.  Each element of the sine triangle is the element above and to the left plus the element that is above and to the right, except from the other table.

For example, find the "-3" in the cosine table.  The number above and to the left is -1, and the number above and to the right, but in the other table, is 2.  Their difference is -3.

In the sine table, find the 4.  It's the sum of the number above and to the left, 3, and the number above and to the right, but in the other table, 1.

Future development ideas for this website

You may notice that the cos table has all even sine powers, so it would be kind of nice to reformat it using sin²x=1-cos²x to eliminate all the sines.  I haven't done that, though.  Maybe another day.

Related Pages in this website

Special Angles

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

Cos of multiples of x, in terms of cos x 

Sin of multiples of x, in terms of cos x

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 the Math Help site is Graeme McRae.
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