Express $\cos 6\theta $ in terms of $\cos \theta$

Question

I think I'm supposed to use the chebyshev polynomials, as in $$ \cos n \theta = T_n(x) = \cos(n \arccos x)$$

But no idea what now?

If you set $y=2\theta$ you can write $\cos 3y$ in terms of $\cos y$ and then $\cos y=\cos 2\theta$ in terms of $\cos \theta$ - the double angle formula you should know, and the triple angle formula is well known, or can be easily derived. — Mark Bennet, Jun 29 '13 at 10:26
Do you know how to express $\cos(x+y)$ in terms of $\cos x, \cos y, \sin x, \sin y$? — Najib Idrissi, Jun 29 '13 at 10:26
The Chebyshev polynomials satisfy
$$T_{n}(x)=\cos (n\arccos x)$$ or

$$T_{n}(\cos \theta )=\cos n\theta ,\quad \theta =\arccos x.$$ — Américo Tavares, Jun 29 '13 at 11:12
How do you expand in this form? Express cos(nx) in terms of cos(n-1)x https://i.sstatic.net/x4MGe.png — N S, Mar 08 '15 at 08:42
http://math.stackexchange.com/questions/1180547/express-cos-nx-in-terms-of-sum-cos-n-1-x If you can help me out here — N S, Mar 08 '15 at 08:48

Américo Tavares · Accepted Answer · 2013-06-29T15:01:20.167

Since $$\cos 2a =2\cos ^{2}a -1, \qquad\sin 2a =2\sin a\cos a,$$ $$\cos (a+b)=\cos a\cos b-\sin a\sin b,$$ and $$ \begin{eqnarray*} \cos 3\theta &=&\cos (2\theta +\theta ) \\ &=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\ &=&( 2\cos ^{2}\theta -1) \cos \theta -2\sin ^{2}\theta \cos \theta \\ &=&( 2\cos ^{2}\theta -1) \cos \theta -2( 1-\cos ^{2}\theta ) \cos \theta \\ &=&4\cos ^{3}\theta -3\cos \theta , \end{eqnarray*} $$ we have $$ \begin{eqnarray*} \cos 6\theta &=&\cos \left( 2\times 3\theta \right) \\ &=&2\cos ^{2}3\theta -1 \\ &=&2( 4\cos ^{3}\theta -3\cos \theta ) ^{2}-1 \\ &=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1. \end{eqnarray*} $$

ADDED: From the definition of the Chebyshev polynomials $$ \begin{equation*} T_{n}(x)=\cos (n\arccos x)\Leftrightarrow T_{n}(\cos \theta )=\cos n\theta ,\quad \theta =\arccos x, \end{equation*} $$

we get $$ \begin{eqnarray*} T_{1}(x) &=&\cos (\arccos x)=x \\ T_{2}(x) &=&\cos (2\arccos x)=2x^{2}-1. \end{eqnarray*} $$

Since they satisfy the recurrence $$ \begin{equation*} T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x), \end{equation*} $$

we have

$$ \begin{eqnarray*} T_{3}(x) &=&2xT_{2}(x)-T_{1}(x) \\ &=&2x( 2x^{2}-1) -x \\ &=&4x^{3}-3x \\ && \\ T_{4}(x) &=&2xT_{3}(x)-T_{2}(x) \\ &=&2x( 4x^{3}-3x) -( 2x^{2}-1) \\ &=&8x^{4}-8x^{2}+1 \\ && \\ T_{5}(x) &=&2xT_{4}(x)-T_{3}(x) \\ &=&2x( 8x^{4}-8x^{2}+1) -( 4x^{3}-3x) \\ &=&16x^{5}-20x^{3}+5x \\ && \\ T_{6}(x) &=&2xT_{5}(x)-T_{4}(x) \\ &=&2x( 16x^{5}-20x^{3}+5x) -( 8x^{4}-8x^{2}+1) \\ &=&32x^{6}-48x^{4}+18x^{2}-1. \end{eqnarray*} $$

Therefore $$ \begin{eqnarray*} \cos 6\theta &=&T_{6}(\cos \theta ) \\ &=&32\left( \cos \theta \right) ^{6}-48\left( \cos \theta \right) ^{4}+18\left( \cos \theta \right) ^{2}-1 \\ &=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1, \end{eqnarray*} $$

as above.

score 2 · Answer 2 · answered Jun 29 '13 at 10:43

2

HINT: Note that, $$\cos 6\theta+i\sin 6\theta=e^{i6\theta }=\left(e^{i\theta }\right)^6=(\cos\theta+i\sin\theta)^6$$ Now expand the RHS using binomial theorem and equate real and imaginary parts.

answered Jun 29 '13 at 10:43

pritam

10,445

wendy.krieger · Answer 3 · 2013-06-29T11:08:27.793

The trick here is that chebyshev polynomials follow a relation in the form of this, when the values are doubled (ie $T(n)=2^n t(n)$)

$$t(n+1) = a \cdot t(n) - t(n-1)$$

So you have in base 'a'.

                                 2    (0
                            1    0    (1
                       1    0   -2    (2
                  1    0   -3    0    (3
              1   0   -4    0    2    (4
         1   0   -5    0    5    0    (5
     1   0  -6    0    9    0   -2    (6

The equation then reads of as $a^6 - 6 a^4 + 9a^2 - 2$, and since $a= 2\cos(\theta)$, and the value is $2 \cos(6\theta)$

$$\cos(6\theta) = 32 \cos^6(\theta) - 48\cos^4(\theta) +18 \cos^2(\theta) - 1$$

Express $\cos 6\theta $ in terms of $\cos \theta$

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