Triple Cosine to Triple Sine

$\cos\theta + \cos^{2}\theta + \cos^{3}\theta = 1$

$\cos\theta + \cos^{3}\theta = 1- \cos^{2}\theta =\sin^{2}\theta$

squaring both side,

$\cos^{2}\theta +2\cos^{4}\theta+ \cos^{6}\theta = \sin^{4}\theta$

$[1-\sin^{2}\theta]+2[(1-\sin^{2}\theta]^{2}+[1-\sin^{2}\theta]^{3}=\sin^{4}\theta$

$[1-\sin^{2}\theta]+[2-4\sin^{2}\theta+2\sin^{4}\theta]+[1-3\sin^{2}\theta+3\sin^{4}\theta-\sin^{6}\theta]=\sin^{4}\theta$

$4-8\sin^{2}\theta+4\sin^{4}\theta=\sin^{6}\theta$

Thus, $a+b+c=4-8+4=\boxed{0}$

For simplicity, let $C=\cos{\theta}$ and $S=\sin{\theta}$ .

It is given that: $C^3+C^2+C =1$

$\Rightarrow (C^3+C^2+C)^2 = C^6 + C^4 + C^2 + 2(C^3C^2+C^3C+C^2C)=1^2$

$\Rightarrow C^6 + 2C^5 + 3C^4 + 2C^3 +C^2=1$

$\Rightarrow C^6 + 2C^2 (C^3+C^2+C) + C^4 +C^2=1$

$\Rightarrow C^6 + 2C^2 (1) + C^4 +C^2= C^6 + C^4 +3C^2 = 1$

$\Rightarrow C^6 = 1 - 3C^2-C^4$

Now:

$\sin^6{\theta} = S^6 = (S^2)^3 = (1-C^2)^3 = 1 - 3C^2+3C^4 - C^6$

$\quad \quad = 1 - 3C^2+3C^4 - (1 - 3C^2-C^4) = 4C^4 = 4(1-S^2)^2$

$\quad \quad = 4(1-2S^2+S^4 = 4 - 8\sin^2{\theta}+4\sin^4{\theta}$

$\Rightarrow a+b+c = 4-8+4 = \boxed{0}$

$\cos+\cos^3=\cos(2-\sin^2)=\sin^2$ $(1-\sin^2)(2-\sin^2)^2=\cos^2(2-\sin^2)^2=\sin^4$ $4-\sin^6-8\sin^2+5\sin^4=\sin^4$ $\sin^6=-4+8\sin^2-4\sin^4$ $a=-4, b=8, c=-4$