# How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?

Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.

Here is what I've worked out:

Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.

$$g(x) = (f(x))^6$$

$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$

So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$

However I'm stucked from here on. Thank you in advance!

• What stops you from running the calcs? – user228113 Jan 3 '16 at 9:09
• You can write $(\cos x)^6$ as a sum of cosines of the forms $r\cos nx$ for suitable rational $r$ and integral $n$ – Mark Bennet Jan 3 '16 at 9:14

$$f\left( x \right) =f\left( 0 \right) +{ f\left( 0 \right) }^{ \prime }x+\frac { { f\left( 0 \right) }^{ \prime \prime } }{ 2! } { x }^{ 2 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime } }{ 3! } { x }^{ 3 }+\frac { { f\left( 0 \right) }^{ \prime \prime \prime \prime } }{ 4! } { x }^{ 4 }+...\\ f\left( x \right) =\cos ^{ 6 }{ x } \\ f\left( 0 \right) =1\\ f^{ \prime }\left( x \right) =-6\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime }\left( 0 \right) =-6\cos ^{ 5 }{ 0 } \sin { 0 } =0\\ f^{ \prime \prime }\left( x \right) ={ \left( -6\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \Rightarrow f^{ \prime \prime }\left( 0 \right) =-6\\ f^{ \prime \prime \prime }\left( x \right) ={ \left( 30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x } } \right) }^{ \prime }=-120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime \prime \prime }\left( 0 \right) =0\\ f^{ \prime \prime \prime \prime }\left( x \right) ={ \left( -120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \right) }^{ \prime }=360\cos ^{ 2 }{ x } \sin ^{ 4 }{ x } -360\sin ^{ 2 }{ x } \cos ^{ 4 }{ x } -300\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +60\cos ^{ 6 }{ x } -180\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +36\cos ^{ 5 }{ x } \Rightarrow \\ f^{ \prime \prime \prime \prime }\left( 0 \right) =96\\ \cos ^{ 6 }{ x } =1-\frac { 6 }{ 2! } { x }^{ 2 }+\frac { 96 }{ 4! } { x }^{ 4 }+...=1-3{ x }^{ 2 }+4{ x }^{ 4 }+..\\$$

I think the simplest way is by expanding to get terms with degrees below 4 we have to choose which combination of terms of original cos(x) series must be chosen and sum up the posibble combination here is the answer:

$$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$ $$=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times...\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)$$ so there is this combinations:

1-$$choosing \; all \; 1's \; from \; 6 \; parentheses \;that\;can \;be \;done \;in \;one \;uniqe \;way\; so \; term \;1 \;is \;produced :\;1$$ 2-$$choosing \; 1's \; from \; 5 \; parentheses \; and- \frac{1}{2}x^2 \; or \; \frac{1}{24}x^4 from \;other \;parentheses \;which \;can \;be \;done \;in\; 6 \;defferent \;ways\; so \; terms \; - \frac{6}{2}x^2 and \frac{6}{24}x^4\;are \;produced :\;- 3\times x^2 + \frac{1}{4}x^4$$

3-$$choosing \; 1's \; from \; 4 \; parentheses \; and- \frac{1}{2}x^2 \; from\; 4\;other \;parentheses \;which \;can \;be \;done \;in\; \binom{6}{2} \;defferent \;ways\; so \; term \; \frac{15}{4}x^4 is \;produced :\;+ \frac{15}{4}x^4$$

4-there is no other term possible with degree below 6. summing up above terms results in the truncated series as below:

$$1 + 3\times x^2 + 4\times x^4 + O(x^6)$$

Let $A=\cos x.$ We have $\cos 2x=2A^2-1.\quad$$\cos 3 x= 4A^3-3A.\quad$$ \cos 4x=2\cos^22x-1=8A^4-8A^2+1.\quad$$\cos 6x=2(\cos 3x)^2-1=32A^6-48A^4+18A^2-1. From the expressions for \cos 2x,\; \cos 4x,\;\cos 6x we have 32A^6=\cos 6x+6\cos 4x+15\cos 2x+10. So \cos^6x=\sum_{n=0}^{\infty}(-1)^nx ^{2n}2^{-5}(6^{2n}+6.4^{2n}+15.2^{2n}+10)/(2n)!.$$\cos x=\frac {e^{ix}+e^{-ix}}2$$so$$32\cos^6x=\cos 6x+6\cos 4x+15\cos 2x+10==(1-18x^2+54x^4+\dots)+(6-48x^2+64x^4-\dots)+(15-30x^2+10x^4-\dots)+10$$Let f(x) = \cos x, and g(x) = f(x)^6 \Rightarrow g'(0) = 6f(0)^5\cdot f'(0), and g'(x) = 6f(x)^5\cdot f'(x) \Rightarrow g''(x) = 30f(x)^4\cdot (f'(x))^2+ 6f(x)^5\cdot f''(x), and continue until you reach the desire exponent when you readily substitute x = 0 into the expression of the g^{(n)}(0), n = 0,1,... If you want to use the Maclaurin series for \cos in order to obtain the first few terms of the Maclaurin series for \cos^6 proceed as follows: Start with$$\cos x=1-{x^2\over2}+{x^4\over24}+?x^6=1-{x^2\over2}\left(1-{x^2\over12}+?x^4\right)\ .$$Using the binomial theorem you then obtain$$\cos^6 x=1-{6\choose 1}{x^2\over2}\left(1-{x^2\over12}+?x^4\right)+{6\choose2}{x^4\over4}(1+?x^2)+?x^6\ .$$Now collect terms:$$\cos^6 x=1-3x^2+4x^4+?x^6\ .$$Since you only need the terms up to and including x^4, it would be easier just to use the Binomial theorem and get$$\cos^6x=\left[1+\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)\right]^6=1+6\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)+15\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)^2+...=1-3x^2+4x^4...$\$