Find the value of $(a+b+c)$ when $\cos\theta+\cos^2\theta+\cos^3\theta=1$ and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$

Question

Given: $\cos\theta+\cos^2\theta+\cos^3\theta=1$ and $\sin^6\theta=a+b\sin^2\theta+c\sin^4\theta$

Then find the value of $(a+b+c)$

Answer 1

Clearly we need to eliminate $\cos\theta$

We have $\displaystyle\cos\theta(1+\cos^2\theta)=1-\cos^2\theta\iff\cos\theta(1+1-\sin^2\theta)=\sin^2\theta$

Squaring we get $$\cos^2\theta(2-\sin^2\theta)^2=(\sin^2\theta)^2$$ $$\iff(1-\sin^2\theta)(2-\sin^2\theta)^2=\sin^4\theta$$

Rearrange to form a six-degree equation in $\sin\theta$ and compare with the given six degree equation

for the constants and the coefficients of $\displaystyle\sin^2\theta,\sin^4\theta,\sin^6\theta$