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)  1cos^{1}x 
cos(2x)  2cos^{2}x – 1 
cos(3x)  4cos^{3}x – 3cos^{1}x 
cos(4x)  8cos^{4}x – 8cos^{2}x + 1 
cos(5x)  16cos^{5}x – 20cos^{3}x + 5cos^{1}x 
cos(6x)  32cos^{6}x – 48cos^{4}x + 18cos^{2}x – 1 
cos(7x)  64cos^{7}x – 112cos^{5}x + 56cos^{3}x – 7cos^{1}x 
cos(8x)  128cos^{8}x – 256cos^{6}x + 160cos^{4}x – 32cos^{2}x + 1 
cos(9x)  256cos^{9}x – 576cos^{7}x + 432cos^{5}x – 120cos^{3}x + 9cos^{1}x 
cos(10x)  512cos^{10}x – 1280cos^{8}x + 1120cos^{6}x – 400cos^{4}x + 50cos^{2}x – 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) = 2n1: 1 3 5 7 9 11 .....The sequence in Row 2 is determined by the formula:
R_1(n) = 2\sum_{i=1}^{n} 2i1: 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_11: 4 20 56 120 220 364The 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_21: 8 48 160 400 840 1568 2688Continue 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  cos^{1}x 
2  2sin x·cos x  2cos^{2}x – 1 = 1 – 2sin^{2}x 
3  3sin x – 4sin^{3}x  4cos^{3}x – 3cos^{1}x 
4  2sin 2x(1 – 2sin^{2}x) = 2sin 2x – 4sin 2x·sin^{2}x  8cos^{4}x – 8cos^{2}x + 1 = 1 – 8sin^{2}x + 8sin^{4}x 
5  5sin^{1}x – 20sin^{3}x + 16sin^{5}x  16cos^{5}x – 20cos^{3}x + 5cos^{1}x 
6  3sin 2x – 4sin^{3}2x  32cos^{6}x – 48cos^{4}x + 18cos^{2}x – 1 = 1 – 18sin^{2}x + 48sin^{4}x – 32sin^{6}x 
7  7sin^{1}x – 56sin^{3}x + 112sin^{5}x – 64sin^{7}x 
64cos^{7}x – 112cos^{5}x + 56cos^{3}x – 7cos^{1}x 
8 

128cos^{8}x – 256cos^{6}x + 160cos^{4}x – 32cos^{2}x + 1 
9 

256cos^{9}x – 576cos^{7}x + 432cos^{5}x – 120cos^{3}x + 9cos^{1}x 
10 

512cos^{10}x – 1280cos^{8}x + 1120cos^{6}x – 400cos^{4}x + 50cos^{2}x  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})^{n1}i\sin{x} + C_{2}^{n}(\cos{x})^{n2}(i\sin{x})^{2} + C_{3}^{n}(\cos{x})^{n3}(i\sin{x})^{3} + C_{4}^{n}(\cos{x})^{n4}(i\sin{x})^{4} + ...Now, group the real part from the imaginary part in expansion.
(ii) [\cos^{n}{x}  C_{2}^{n}(\cos{x})^{n2}(\sin{x})^{2} + C_{4}^{n}(\cos{x})^{n4}(\sin{x})^{4}  C_{6}^{n}(\cos{x})^{n6}(\sin{x})^{6} + ...] +i[C_{1}^{n}(\cos{x})^{n1}\sin{x}  C_{3}^{n}(\cos{x})^{n3}(\sin{x})^{3} + C_{5}^{n}(\cos{x})^{n5}(\sin{x})^{5}  C_{7}^{n}(\cos{x})^{n7}(\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})^{n1}\sin{x}  C_{3}^{n}(\cos{x})^{n3}(\sin{x})^{3} + C_{5}^{n}(\cos{x})^{n5}(\sin{x})^{5}  C_{7}^{n}(\cos{x})^{n7}(\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})^{n2}(\sin{x})^{2} + C_{4}^{n}(\cos{x})^{n4}(\sin{x})^{4}  C_{6}^{n}(\cos{x})^{n6}(\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})^{n2}(1\cos^{2}{x}) + C_{4}^{n}(\cos{x})^{n4}(1\cos^{2}{x})^{2}  C_{6}^{n}(\cos{x})^{n6}(1\cos^{2}{x})^{3} + ...
\cos{(nx)} = \cos^{n}{x}  C_{2}^{n}[(\cos{x})^{n2}  \cos^{n}{x})] + C_{4}^{n}[(\cos{x})^{n4}  C_{1}^{2}\cos{x}^{n2} + \cos^{n}{x}]  C_{6}^{n}[(\cos{x})^{n6}  C_{1}^{3}(\cos{x})^{n4} + C_{2}^{3}(\cos{x})^{n2}  C_{3}^{3}(\cos{x})^{n}] + C_{8}^{n}[(\cos{x})^{n8}  C_{1}^{4}(\cos{x})^{n6} + C_{2}^{4}(\cos{x})^{n4}  C_{3}^{4}(\cos{x})^{n2} + 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}^{n2} + [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}^{n4}  [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}^{n6} + ...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_{i1}^{i} \right )(\cos{x})^{n2} + \left (\sum_{i=2}^{\left \lfloor \frac{n}{2} \right \rfloor  1}C_{2i}^n\cdot C_{i2}^{i} \right )(\cos{x})^{n4}  \left (\sum_{i=3}^{\left \lfloor \frac{n}{2} \right \rfloor  2}C_{2i}^n\cdot C_{i3}^{i} \right )(\cos{x})^{n6} + \left (\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor  3}C_{2i}^n\cdot C_{i4}^{i} \right )(\cos{x})^{n8}  \left (\sum_{i=5}^{\left \lfloor \frac{n}{2} \right \rfloor  4}C_{2i}^n\cdot C_{i5}^{i} \right )(\cos{x})^{n10} + ..., 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_{i1}^{i}=2^{n3}(n) = \frac{2^{n3}}{1!}(n), where n ≥ 2
\sum_{i = 2}^{\left \lfloor \frac{n}{2} \right \rfloor  1}C_{2i}^{n}\cdot C_{i2}^{i}=2^{n6}(n)(n3) = \frac{2^{n5}}{2!}(n)(n3), where n ≥ 4
\sum_{i=3}^{\left \lfloor \frac{n}{2} \right \rfloor  2}C_{2i}^{n}\cdot C_{i3}^{i} = 2^{n6}\left [ \frac{7}{2} + \frac{9}{2}(n7) + \frac{5}{4}(n7)(n8) + \frac{1}{12}(n7)(n8)(n9) \right ] = 2^{n6}\left ( \frac{1}{12} \right )(n)(n4)(n5) = 2^{n8}(1/3)(n)(n4)(n5) = \frac{2^{n7}}{3!}(n)(n4)(n5), where n ≥ 6
\sum_{i=4}^{\left \lfloor \frac{n}{2} \right \rfloor  3}C_{2i}^{n}\cdot C_{i4}^{i} = 2^{n12}\left ( \frac{1}{3} \right )(n)(n5)(n6)(n7) = \frac{2^{n9}}{4!}(n)(n5)(n6)(n7), 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_{i5}^{i} = \frac{2^{n11}}{5!}(n)(n6)(n7)(n8)(n9), where n ≥ 10
Does this compute the coefficients? Let’s start with n = 10.
\frac{2^{1011}}{5!}(10)(106)(107)(108)(109) = \frac{2^{1}}{5\cdot 4!}(10)(4)(3)(2)(1)=1. So far so good. I'll skips the inbetween 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 cos^{10}(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 cos^{8}x 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 cos^{6}x 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 cos^{4}x 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 cos^{2}x coefficient:
C_{8}^{10}C_{0}^{4} + C_{10}^{10}C_{1}^{5} = 45\cdot1 + 1\cdot5 = 45 + 5 = 50
For the cos^{0}x (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)  –1sin^{1}x  –1 
cos(2x)  –4cos^{1}xsin^{1}x  –4 
cos(3x)  –12cos^{2}xsin^{1}x + 3sin^{1}x  –12 + 3 = –9 
cos(4x)  –32cos^{3}xsin^{1}x + 16cos^{1}xsin^{1}x  –32 + 16 = –16 
cos(5x)  –80cos^{4}xsin^{1}x + 60cos^{2}xsin^{1}x – 5sin^{1}x  –80 + 60 – 5 = –25 
cos(6x)  –192cos^{5}xsin^{1}x + 192cos^{3}xsin^{1}x – 36cos^{1}xsin^{1}x  –192 + 192 – 36 = –36 
cos(7x)  –448cos^{6}xsin^{1}x + 560cos^{4}xsin^{1}x – 168cos^{2}xsin^{1}x + 7sin^{1}x  –448 + 560 – 168 + 7 = –49 
cos(8x)  –1024cos^{7}xsin^{1}x + 1536cos^{5}xsin^{1}x – 640cos^{3}xsin^{1}x + 64cos^{1}xsin^{1}x  –1024 + 1536 – 640 + 64 = –64 
cos(9x)  –2304cos^{8}xsin^{1}x + 4032cos^{6}xsin^{1}x – 2160cos^{4}xsin^{1}x + 360cos^{2}xsin^{1}x – 9sin^{1}x  –2304 + 4032 – 2160 + 360 – 9 = –81 
cos(10x)  –5120cos^{9}xsin^{1}x + 10240cos^{7}xsin^{1}x – 6720cos^{5}xsin^{1}x + 1600cos^{3}xsin^{1}x – 100cos^{1}xsin^{1}x  –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_{i4}^{i} \right )\cos{x}^{n8}. 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_{n1}}{1!}i + \frac{K_{n2}}{2!}i(i  1) + \frac{K_{n3}}{3!}i(i  1)(i  2) + \frac{K_{n4}}{4!}i(i  1)(i  2)(i  3) + \frac{K_{n5}}{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 nlevel 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 nlevel differences to gives us the rule for this sequence. So, we have K_{4} = 1, K_{3} = 3.5, K_{2} = 4.5, K_{1} = 2.5, and K_{0} = 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)(n1) + \frac{5}{12} (n)(n1)(n2) + \frac{1}{48}(n)(n1)(n2)(n3)
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)(n1) + \frac{5}{12}(n)(n1)(n2) + \frac{1}{48}(n)(n1)(n2)(n3)\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_{i4}^{i} \right ) = 2^{n8}\left [1 + \frac{7}{2}(n8) + \frac{9}{4}(n8)(n9) + \frac{5}{12}(n8)(n9)(n10) + \frac{1}{48}(n8)(n9)(n10)(n11)\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_{i4}^{i} \right ) = 2^{n8}\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_{i4}^{i} \right ) = \frac{2^{n12}}{3}(n)(n5)(n6)(n7).