Sine of multiples of x, in terms of cos x
As you know from Trig Equiv and Sin or Cos
3x,
sin 2x = 2 sin x cos x,
sin 3x = -sin3x + 3cos2x sin x
If you keep going, you'll see the sines of all the multiples of x have only
odd powers of sin x, so you can always use
sin2x=(1-cos2x) to express sin nx in terms sin x times an
expression containing cos x:
sin 2x = (sin x) (2 cos x)
sin 3x = (sin x) (-1 + 4cos2x)
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
sin y+x = sin y cos x + cos y sin x, and
sin y-x = sin y cos x - cos y sin x
So the sum of sin y+x and sin y-x is 2 sin y cos x. So we get
this useful trig equivalence:
sin y+x = 2 sin y cos x - sin 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:
sin nx = 2 sin[(n-1)x] cos x - sin[(n-2)x]
Now that we know cos 2x and cos 3x, let's see how to use this fact to find
cos 4x:
sin 2x = (sin x) (2 cos x)
sin 3x = (sin x) (-1 + 4cos2x)
sin 4x = 2 sin 3x cos x - sin 2x
sin 4x = (sin x) ( 2 (-1 + 4cos2x)(cos x) - (2 cos x) )
sin 4x = (sin x) ( (-2 cos x + 8cos3x) - (2 cos x) )
sin 4x = (sin x) (-4 cos x + 8cos3x)
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:
sin 1x = (sin) (1)
sin 2x = (sin) (2cos)
sin 3x = (sin) (-1+4cos2)
sin 4x = (sin) (-4cos+8cos3)
sin 5x = (sin) (1-12cos2+16cos4)
sin 6x = (sin) (6cos-32cos3+32cos5)
sin 7x = (sin) (-1+24cos2-80cos4+64cos6)
sin 8x = (sin) (-8cos+80cos3-192cos5+128cos7)
sin 9x = (sin) (1-40cos2+240cos4-448cos6+256cos8)
sin 10x = (sin) (10cos-160cos3+672cos5-1024cos7+512cos9)
sin 11x = (sin) (-1+60cos2-560cos4+1792cos6-2304cos8+1024cos10)
sin 12x = (sin) (-12cos+280cos3-1792cos5+4608cos7-5120cos9+2048cos11)
sin 13x = (sin) (1-84cos2+1120cos4-5376cos6+11520cos8-11264cos10+4096cos12)
sin 14x = (sin) (14cos-448cos3+4032cos5-15360cos7+28160cos9-24576cos11+8192cos13)
sin 15x = (sin) (-1+112cos2-2016cos4+13440cos6-42240cos8+67584cos10-53248cos12+16384cos14)
sin 16x = (sin) (-16cos+672cos3-8064cos5+42240cos7-112640cos9+159744cos11-114688cos13+32768cos15)
sin 17x = (sin) (1-144cos2+3360cos4-29568cos6+126720cos8-292864cos10+372736cos12-245760cos14+65536cos16)
sin 18x = (sin) (18cos-960cos3+14784cos5-101376cos7+366080cos9-745472cos11+860160cos13-524288cos15+131072cos17)
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:
sin nx = 2 sin[(n-1)x] cos x - sin[(n-2)x]
I'll give you the first few rows of the triangle:
sin nx
1
2
-1 4
-4 8
1 -12 16
6 -32 32
-1 24 -80 64
-8 80 -192 128
1 -40 240 -448 256
10 -160 672 -1024 512
-1 60 -560 1792 -2304 1024
-12 280 -1792 4608 -5120 2048
1 -84 1120 -5376 11520 -11264 4096
14 -448 4032 -15360 28160 -24576 8192
-1 112 -2016 13440 -42240 67584 -53248 16384
-16 672 -8064 42240 -112640 159744 -114688 32768
1 -144 3360 -29568 126720 -292864 372736 -245760 65536
18 -960 14784 -101376 366080 -745472 860160 -524288 131072
Related Pages in this website
Special Angles
Trig Equivalences
Sin or Cos 3x, 4x, etc. -- trig
functions of any multiple of an angle.
Cos 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.
