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 Math Help > Trigonometry > Trig Equivalences > Sin or Cos 3x, 4x, etc.

# 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

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

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.