Multiple Angle Identities for Cosine and Sine

C o n c e p t s

This topic requires familiarity with the following concepts:

  • Conic equations
  • Differentiation
  • Integration
  • Rotating graphs
  • Golden ratio

We know the identities for sin(2x) and cos(2x), but what about sin(nx) and cos(nx)? I have found a method to find the identities for cos(nx) and other properties relating sin(nx) and cos(nx) in the form of \sum_{i}^{ }a_i\cos ^i(x). First we find the identities for cos(3x) to cos(10x) exclusively in terms of cosx using the sum formula of cosine. They are listed here:

Table 1: Cosine Identities for n = 1 to 10

cos(1x) 1cos1x
cos(2x) 2cos2x – 1
cos(3x) 4cos3x – 3cos1x
cos(4x) 8cos4x – 8cos2x + 1
cos(5x) 16cos5x – 20cos3x + 5cos1x
cos(6x) 32cos6x – 48cos4x + 18cos2x – 1
cos(7x) 64cos7x – 112cos5x + 56cos3x – 7cos1x
cos(8x) 128cos8x – 256cos6x + 160cos4x – 32cos2x + 1
cos(9x) 256cos9x – 576cos7x + 432cos5x – 120cos3x + 9cos1x
cos(10x) 512cos10x – 1280cos8x + 1120cos6x – 400cos4x + 50cos2x – 1

Notice that by increasing the n in cos(nx), the multiple of angle x, the number of terms increase, and we have a trigonometric polynomial each time. We'll define a trigonometric polynomial as a polynomial containing either cosine or sine terms which have a 1 as the multiple of the angle x: \sum_{i}^{ }a_i\cos ^i(x). The trigonometric polynomial identities of cosine, where n is the multiple of angle x, have the following special characteristics:

  • The coefficient of the term of the greatest degree is a power of 2.
  • The signs alternate from positive to negative when the terms are arranged in decreasing degrees.
  • The degrees of the terms decrease by 2. If the multiple is even, the degrees of all the terms are even. If the multiple is odd, the degrees of all the terms are odd.
  • The greatest degree of the trigonometric polynomial identity is the multiple of angle x.
  • The sum of the coefficients of the trigonometric polynomial identity is always 1. This can be proved easily by noting that cos(0) = 1.

We only need to find a way to generate the coefficients of the terms other than the greatest degree term. I have found one method shown here.

First, write the odd numbers as the first row starting with 1:

R_1(n) = 2n-1: 1 3 5 7 9 11 .....

The sequence in Row 2 is determined by the formula:

R_1(n) = 2\sum_{i=1}^{n} 2i-1: 2 8 18 32 50 72 .....

The sequence in Row 3 is determined by the formula:

R_1(n) = 4\sum_{i=1}^{n}\sum_{i_1=1}^{i} 2i_1-1: 4 20 56 120 220 364

The pattern is evident and the formula for Row 4 is:

R_1(n) = 4\sum_{i=1}^{n}\sum_{i_1=1}^{i}\sum_{i_2=1}^{i_1} 2i_2-1: 8 48 160 400 840 1568 2688

Continue this process to find as many rows as desired.

Row 1: 1 3 5 7 9 11 ...
Row 2: 2 8 18 32 50 72 ...
Row 3: 4 20 56 120 220 364 ...
Row 4: 8 48 160 400 840 1568 ...
Row 5: 16 112 432 1232 2912 6048 ...
Row 6: 32 256 1120 3584 9408 21504 ...

Column 1 contains coefficient of the highest degree terms. Column 2 contains coefficients of the next highest degree terms, and so on. But the columns after Column 1 have to be dropped down. Drop the columns down as shown below.

Row 1: 1 0 0 0 0 0 ...
Row 2: 2 1 0 0 0 0 ...
Row 3: 4 3 0 0 0 0 ...
Row 4: 8 8 1 0 0 0 ...
Row 5: 16 20 5 0 0 0 ...
Row 6: 32 48 18 1 0 0 ...
Row 7: 64 112 56 7 0 0 ...
Row 8: 128 256 160 32 1 0 ...

Column 2 should start with one 0 and a 1. Column 3 should start with three 0's and a 1. Column 4 should start with five 0's and a 1, and so on. Each time an odd number of 0's are added and a 1 is added because for cos(2nx) where 2n is an even number, our trigonometric polynomial ends with a positive or a negative 1. The signs of the trigonometric polynomial can be determined easily because they alternate. Finally, the above table of rows and columns represent the coefficients of the trigonometric polynomial for cos(nx). (Row n represents the coefficients the trigonometric polynomial of cos(nx).) All these identities can be confirmed with a graphing utility.

We've only examined the identities for cos(nx). The identities of sin(nx) are unusual. Because sin(2x) equals 2sin x·cos x, sin(2nx) where 2n is even cannot be expressed exclusively in terms of sin x and they do not form a trigonometric polynomial according to our definition.

Table 2: Cosine and Sine Identities Comparison

N SIN(NX) COS(NX)
1 sin x cos1x
2 2sin x·cos x 2cos2x – 1 = 1 – 2sin2x
3 3sin x – 4sin3x 4cos3x – 3cos1x
4 2sin 2x(1 – 2sin2x) = 2sin 2x – 4sin 2x·sin2x 8cos4x – 8cos2x + 1 = 1 – 8sin2x + 8sin4x
5 5sin1x – 20sin3x + 16sin5x 16cos5x – 20cos3x + 5cos1x
6 3sin 2x – 4sin32x 32cos6x – 48cos4x + 18cos2x – 1 = 1 – 18sin2x + 48sin4x – 32sin6x
7 7sin1x

– 56sin3x + 112sin5x – 64sin7x

64cos7x – 112cos5x + 56cos3x – 7cos1x
8

 

128cos8x – 256cos6x + 160cos4x – 32cos2x + 1
9

 

256cos9x – 576cos7x + 432cos5x – 120cos3x + 9cos1x
10

 

512cos10x – 1280cos8x + 1120cos6x – 400cos4x + 50cos2x - 1

The Fibonacci Sequence

The Fibonacci sequence has a tendency to appear in the strangest places as it does in Pascal's triangle. Our trigonometric polynomial is no exception for the Fibonacci sequence to appear. Let's study the coefficient more closely for cos(nx).

Add up the terms going diagonally from left to right to reveal the sequence: 1, 1, 2, 3, 5, 8, 13, 21....

Proof of the General Trigonometric Polynomial Identity

The method for finding the identity of cos(nx) shown above is not practical for large values of n. There is a general formula for finding the coefficients of the identities. Here, I will reveal the general trigonometric polynomial identity for cos(nx) in a proof which only involves algebra.

Euler proved that e^{xi} = \cos(x) + i\sin(x), where i is the imaginary unit. Using this equation we will find the general identity for cos(nx).

(e^{xi})^{n} = (\cos(x) + i\sin(x))^{n}

Expanding (\cos(x) + i\sin(x))^{n}, we have:

(i) (e^{xi})^{n} = (\cos(x) + i\sin(x))^{n} = \cos^{n}{x} + C_{1}^{n}(\cos{x})^{n-1}i\sin{x} + C_{2}^{n}(\cos{x})^{n-2}(i\sin{x})^{2} + C_{3}^{n}(\cos{x})^{n-3}(i\sin{x})^{3} + C_{4}^{n}(\cos{x})^{n-4}(i\sin{x})^{4} + ...

Now, group the real part from the imaginary part in expansion.

(ii) [\cos^{n}{x} - C_{2}^{n}(\cos{x})^{n-2}(\sin{x})^{2} + C_{4}^{n}(\cos{x})^{n-4}(\sin{x})^{4} - C_{6}^{n}(\cos{x})^{n-6}(\sin{x})^{6} + ...] +i[C_{1}^{n}(\cos{x})^{n-1}\sin{x} - C_{3}^{n}(\cos{x})^{n-3}(\sin{x})^{3} + C_{5}^{n}(\cos{x})^{n-5}(\sin{x})^{5} - C_{7}^{n}(\cos{x})^{n-7}(\sin{x})^{7} + ...]

But since (e^{xi})^{n} = e^{nxi} = \cos{(nx)} + i\sin{(nx)}, which equals the expansion above, the real part in the expansion must be cos(nx), and the imaginary part must be isin(nx). Therefore, we have the following identity of sine:

\sin{(nx)} = C_{1}^{n}(\cos{x})^{n-1}\sin{x} - C_{3}^{n}(\cos{x})^{n-3}(\sin{x})^{3} + C_{5}^{n}(\cos{x})^{n-5}(\sin{x})^{5} - C_{7}^{n}(\cos{x})^{n-7}(\sin{x})^{7} ...

This expression is difficult to simplify but the next property for cosine is fairly easier:

\cos{(nx)} = \cos^{n}{x} - C_{2}^{n}(\cos{x})^{n-2}(\sin{x})^{2} + C_{4}^{n}(\cos{x})^{n-4}(\sin{x})^{4} - C_{6}^{n}(\cos{x})^{n-6}(\sin{x})^{6} + ...

Because \cos^{2}{x} = 1 - \sin^{2}{x}, we can substitute this in the above relationship to rewrite the relationship completely in terms of cosx.

\cos{(nx)} = \cos^{n}{x} - C_{2}^{n}(\cos{x})^{n-2}(1-\cos^{2}{x}) + C_{4}^{n}(\cos{x})^{n-4}(1-\cos^{2}{x})^{2} - C_{6}^{n}(\cos{x})^{n-6}(1-\cos^{2}{x})^{3} + ...

\cos{(nx)} = \cos^{n}{x} - C_{2}^{n}[(\cos{x})^{n-2} - \cos^{n}{x})] + C_{4}^{n}[(\cos{x})^{n-4} - C_{1}^{2}\cos{x}^{n-2} + \cos^{n}{x}] - C_{6}^{n}[(\cos{x})^{n-6} - C_{1}^{3}(\cos{x})^{n-4} + C_{2}^{3}(\cos{x})^{n-2} - C_{3}^{3}(\cos{x})^{n}] + C_{8}^{n}[(\cos{x})^{n-8} - C_{1}^{4}(\cos{x})^{n-6} + C_{2}^{4}(\cos{x})^{n-4} - C_{3}^{4}(\cos{x})^{n-2} + C_{4}^{4}(\cos{x})^{n}]...

Collecting the like terms gives us the following Cosine Identity:

\cos{(nx)} = [1 + C_{2}^{n} + C_{4}^{n} + C_{6}^{n} + C_{8}^{n} + ....]\cos^{n}(x) - [C_{2}^{n}C_{0}^{1} + C_{4}^{n}C_{1}^{2} + C_{6}^{n}C_{2}^{3} + C_{8}^{n}C_{3}^{4} + ...]\cos{x}^{n-2} + [C_{4}^{n}C_{0}^{2} + C_{6}^{n}C_{1}^{3} + C_{8}^{n}C_{2}^{4} + C_{10}^{n}C_{3}^{5} + ...]\cos{x}^{n-4} - [C_{6}^{n}C_{0}^{3} + C_{8}^{n}C_{1}^{4} + C_{10}^{n}C_{2}^{5} + C_{12}^{n}C_{3}^{6} + ...]\cos{x}^{n-6} + ...

From (ii), we have this nice multiple angle identity for cosine.

The Multiple Angle Cosine Identity

\cos{(nx)}=\left (\sum_{i=0}^{\left \lfloor \frac{n}{2} \right \rfloor+1}C_{2i}^n \right )(\cos{x})^{n} - \left (\sum_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}C_{2i}^n\cdot C_{i-1}^{i} \right )(\cos{x})^{n-2} + \left (\sum_{i=2}^{\left \lfloor \frac{n}{2} \right \rfloor - 1}C_{2i}^n\cdot C_{i-2}^{i} \right )(\cos{x})^{n-4} - \left (\sum_{i=3}^{\left \lfloor \frac{n}{2} \right \rfloor - 2}C_{2i}^n\cdot C_{i-3}^{i} \right )(\cos{x})^{n-6} + \left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^n\cdot C_{i-4}^{i} \right )(\cos{x})^{n-8} - \left (\sum_{i=5}^{\left \lfloor \frac{n}{2} \right \rfloor - 4}C_{2i}^n\cdot C_{i-5}^{i} \right )(\cos{x})^{n-10} + ..., where \left \lfloor \frac{n}{2} \right \rfloor is the floor function.

The first 5 coefficients are can be reduced to the following general formulas:

\sum_{i = 0}^{\left \lfloor \frac{n}{2} \right \rfloor + 1 }C_{2i}^{n}=2^n, where n ≥ 0

\sum_{i = 1}^{\left \lfloor \frac{n}{2} \right \rfloor }C_{2i}^{n}\cdot C_{i-1}^{i}=2^{n-3}(n) = \frac{2^{n-3}}{1!}(n), where n ≥ 2

\sum_{i = 2}^{\left \lfloor \frac{n}{2} \right \rfloor - 1}C_{2i}^{n}\cdot C_{i-2}^{i}=2^{n-6}(n)(n-3) = \frac{2^{n-5}}{2!}(n)(n-3), where n ≥ 4

\sum_{i=3}^{\left \lfloor \frac{n}{2} \right \rfloor - 2}C_{2i}^{n}\cdot C_{i-3}^{i} = 2^{n-6}\left [ \frac{7}{2} + \frac{9}{2}(n-7) + \frac{5}{4}(n-7)(n-8) + \frac{1}{12}(n-7)(n-8)(n-9) \right ] = 2^{n-6}\left ( \frac{1}{12} \right )(n)(n-4)(n-5) = 2^{n-8}(1/3)(n)(n-4)(n-5) = \frac{2^{n-7}}{3!}(n)(n-4)(n-5), where n ≥ 6

\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^{n}\cdot C_{i-4}^{i} = 2^{n-12}\left ( \frac{1}{3} \right )(n)(n-5)(n-6)(n-7) = \frac{2^{n-9}}{4!}(n)(n-5)(n-6)(n-7), where n ≥ 8

However, it seems finding the general formula for other powers seems more and more complex when done manually. I have shown how to derive the formula for the 6th term below. However, a pattern does seem to emerge (hence I’ve written the formulas with the factorials to see if a pattern emerges.) Assuming this pattern is correct, let’s write the formula for the 7th term:

\sum_{i=5}^{\left \lfloor \frac{n}{2} \right \rfloor - 4}C_{2i}^{n}\cdot C_{i-5}^{i} = \frac{2^{n-11}}{5!}(n)(n-6)(n-7)(n-8)(n-9), where n ≥ 10

Does this compute the coefficients? Let’s start with n = 10.

\frac{2^{10-11}}{5!}(10)(10-6)(10-7)(10-8)(10-9) = \frac{2^{-1}}{5\cdot 4!}(10)(4)(3)(2)(1)=1. So far so good. I'll skips the in-between steps going forward.

\frac{2^{0}}{5!}(11)(5)(4)(3)(2) = 11. Works again. Let’s continue with n = 12.

\frac{2^{1}}{5!}(12)(6)(5)(4)(3) = 72. Doing great so far. And now, n = 13.

\frac{2^{2}}{5!}(13)(7)(6)(5)(4) = 364 Amazing! The pattern holds up. This is not a rigorous proof required by mathematics. I’ll leave that up to others to figure that out.

Manual Calculation for cos(10x)

For cos10(x) coefficient:

C_{0}^{10}+C_{2}^{10}+C_{4}^{10}+C_{6}^{10}+C_{8}^{10}+C_{10}^{10} = 1 + 45 + 210 + 210 + 45 + 1 = 512

For the cos8x coefficient:

C_{2}^{10}C_{0}^{1} + C_{4}^{10}C_{1}^{2} + C_{6}^{10}C_{2}^{3} + C_{8}^{10}C_{3}^{4}+C_{10}^{10}C_{4}^{5} = 45\cdot1 + 210\cdot2 + 210\cdot3 + 45\cdot4 + 1\cdot5 = 45 + 420 + 630 + 180 +5 = 1280

For the cos6x coefficient:

C_{4}^{10}C_{0}^{2} + C_{6}^{10}C_{1}^{3} + C_{8}^{10}C_{2}^{4} + C_{10}^{10}C_{3}^{5} = 210\cdot1 + 210\cdot3 + 45\cdot6 + 1\cdot10 = 210 + 630 + 270 + 10 = 1120

For the cos4x coefficient:

C_{6}^{10}C_{0}^{3} + C_{8}^{1}0C_{1}^{4} + C_{10}^{10}C_{2}^{5} = 210\cdot1 + 45\cdot4 + 1\cdot10 = 210 + 180 + 10 = 400

For the cos2x coefficient:

C_{8}^{10}C_{0}^{4} + C_{10}^{10}C_{1}^{5} = 45\cdot1 + 1\cdot5 = 45 + 5 = 50

For the cos0x (the constant):

C_{10}^{10}C_{0}^{5} = 1\cdot1 = 1

Pattern of Squares

Interesting series appears when obtaining the derivatives of the identities and adding their coefficients.

Derivative Sum of Coefficients
cos(1x) –1sin1x –1
cos(2x) –4cos1xsin1x –4
cos(3x) –12cos2xsin1x + 3sin1x –12 + 3 = –9
cos(4x) –32cos3xsin1x + 16cos1xsin1x –32 + 16 = –16
cos(5x) –80cos4xsin1x + 60cos2xsin1x – 5sin1x –80 + 60 – 5 = –25
cos(6x) –192cos5xsin1x + 192cos3xsin1x – 36cos1xsin1x –192 + 192 – 36 = –36
cos(7x) –448cos6xsin1x + 560cos4xsin1x – 168cos2xsin1x + 7sin1x –448 + 560 – 168 + 7 = –49
cos(8x) –1024cos7xsin1x + 1536cos5xsin1x – 640cos3xsin1x + 64cos1xsin1x –1024 + 1536 – 640 + 64 = –64
cos(9x) –2304cos8xsin1x + 4032cos6xsin1x – 2160cos4xsin1x + 360cos2xsin1x – 9sin1x –2304 + 4032 – 2160 + 360 – 9 = –81
cos(10x) –5120cos9xsin1x + 10240cos7xsin1x – 6720cos5xsin1x + 1600cos3xsin1x – 100cos1xsin1x –5120 + 10240 – 6720 + 1600 = 100 = –100

The General Formula for the Coefficients

Finding the coefficients seems like a lot of work. However, the first 5 general formulas are given above. Here, I’ll show a way to find these coefficients using the 5th term.

The fifth term is \left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^{n}\cdot C_{i-4}^{i} \right )\cos{x}^{n-8}. From the summation method I showed above and using Excel, the sequence this produces is 1, 9, 50, 220, 840, 2912, 9408, 26880, ….

Now, to find the rule for this sequence, we will have to use my general formula for the polynomial series, which is given by:

f(i) = K_{n} + \frac {K_{n-1}}{1!}i + \frac{K_{n-2}}{2!}i(i - 1) + \frac{K_{n-3}}{3!}i(i - 1)(i - 2) + \frac{K_{n-4}}{4!}i(i - 1)(i - 2)(i - 3) + \frac{K_{n-5}}{5!}i(i - 1)(i - 2)(i - 3)(i - 4) + ...

Knowing that all the coefficients have a power of 2, we will have to divide out the power of 2, then find the n-level differences. Below, this has been done using Excel:

1 9 50 220 840 2912 9408 28800 84480
Divide by 2n 1 4.5 12.5 27.5 52.5 91 147 225 330
1st level diff 3.5 8 15 25 38.5 56 78 105 137.5
2nd level diff 4.5 7 10 13.5 17.5 22 27 32.5 38.5
3rd level diff 2.5 3 3.5 4 4.5 5 5.5 6 6.5
4th level diff 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

The 4th level differences are constant at 0.5. Hence, we expect this sequence to be a 4th degree sequence. We are only interested in the first terms of the n-level differences to gives us the rule for this sequence. So, we have K4 = 1, K3 = 3.5, K2 = 4.5, K1 = 2.5, and K0 = 0.5. These are the only numbers we are interested in to find the general formula. The polynomial is a 4th degree polynomial.

Plugging these values in gives us:

f(n) = 1+\frac{7}{2}(n) + \frac{9}{4}(n)(n-1) + \frac{5}{12} (n)(n-1)(n-2) + \frac{1}{48}(n)(n-1)(n-2)(n-3)

When n = 0, f(0) = 1. Hence, the power of n that corresponds to this is 2n. So the formula for our coefficients is:

f(n) = 2^n\left [1 + \frac{7}{2}(n) + \frac{9}{4}(n)(n-1) + \frac{5}{12}(n)(n-1)(n-2) + \frac{1}{48}(n)(n-1)(n-2)(n-3)\right ]

However, these coefficients only come into play when n = 8. So we have to shift these by 8.

So the sequence rule is as follows:

\left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^{n}\cdot C_{i-4}^{i} \right ) = 2^{n-8}\left [1 + \frac{7}{2}(n-8) + \frac{9}{4}(n-8)(n-9) + \frac{5}{12}(n-8)(n-9)(n-10) + \frac{1}{48}(n-8)(n-9)(n-10)(n-11)\right ]

The formulas for the other coefficients reduces nicely. Let’s multiply this out to see if the same happens to this formula.

Using a computer, this expands nicely to:

\left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^{n}\cdot C_{i-4}^{i} \right ) = 2^{n-8}\left ( \frac{n^4}{48} - \frac{3n^3}{8} + \frac{107n^2}{48} - \frac{35n}{8} \right ), which itself factors even more nicely as:

\left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor - 3}C_{2i}^{n}\cdot C_{i-4}^{i} \right ) = \frac{2^{n-12}}{3}(n)(n-5)(n-6)(n-7).