<s>[INST] How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem? How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$
using the tensor product and the Clebsch Gordan Theorem?
 [/INST] We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left( 
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left( 
\theta \right)  \right) ^{6}
$$
expanding the right-Hand side we get
$$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left( 
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left( 
\theta \right)  \right) ^{6}
$$
this is the same!
$$\cos(2x)=2\cos(x)^2-1$$
$$\cos(4x)=8\cos(x)^4-8\cos(x)^2+1$$
$$\cos(6x)=32\cos(x)^6-48\cos(x)^4+18\cos(x)^2-1$$
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