Question

Let n be an odd natural number. Prove that there exist polynomials pn(z),qn(z) such that for all real x,
pn(cosx)=cos(nx)
qn(sinx)=sin(nx)

here pn and qn means p sub n & q sub n

Answer 1

Notice
\[\cos (3 x)=\cos ^3(x)-3 \sin ^2(x) \cos (x)=\\
\cos ^3(x)-3 \cos (x)
\left(1-\cos ^2(x)\right)=\\4 \cos ^3(x)-3 \cos (x)

\]

Answer 2

The general case can be done by induction

Answer 3

You will have to do the sin and cos separately

Answer 4

You also can do it by using complex powers e
\[\Large
\cos(nx) =\frac { e^{i n x} +e^{-i n x} }{2}=\\
\Large
\frac { \left(e^{i x} \right)^n+\left(e^{-i x} \right)^n}{2}
\]

Answer 5

\[\Large
\cos(nx) =\frac { e^{i n x} +e^{-i n x} }{2}=\\
\Large
\frac { \left(\cos( x) + i \sin( x) \right)^n+\left(\cos(x) -i \sin(x) \right)^n}{2}=
\Large

\]

Answer 6

Use after the binomial Theorem
See this for details
http://mathworld.wolfram.com/Multiple-AngleFormulas.html

Answer 7

are there are no conditions on p_n (cosx) such that it is an odd polynomial?

Answer 8

Here is the case for n=11
\[\Large
\cos (11 x)=1024 \cos ^{11}(x)-2816 \cos ^9(x)+\\
\Large
2816 \cos ^7(x)-1232 \cos
^5(x)+220 \cos ^3(x)-11 \cos (x)

\]

Answer 9

n=12
\[
\cos (12 x)=2048 \cos ^{12}(x)-6144 \cos ^{10}(x)+\\6912 \cos ^8(x)-3584
\cos ^6(x)+840 \cos ^4(x)-72 \cos ^2(x)+1
\]

Answer 10

It will be a polynomial with degree n

Answer 11

If you want to prove it by induction, use Chebyshev's recurrence formulae
\[
\Large
\cos (n x)=2 \cos (x) \cos ((n-1) x)-\cos ((n-2) x)\\

\Large
\sin (n x)=2 \cos (x) \sin ((n-1) x)-\sin ((n-2) x)\\


\]
@experimentX

Answer 12

\[

\sin (11 x)=-1024 \sin ^{11}(x)+\\2816 \sin ^9(x)-2816 \sin ^7(x)+\\1232 \sin
^5(x)-220 \sin ^3(x)+11 \sin (x)
\]

Answer 13

I believe, you only need n odd for the sin part

Answer 14

@satellite73