Question

Sin A - 2 sin cube A --------------------------- 2 cos cube A - cos A

Answer 1

$\frac{sin\alpha-2sin^3\alpha}{2cos^3\alpha-cos\alpha}=\frac{sin\alpha(1-2sin^2\alpha)}{cos\alpha(2cos^2\alpha-1)}=\frac{sin\alpha(1-sin^2\alpha-sin^2\alpha)}{cos\alpha(cos^2\alpha+cos^2\alpha-1)}\\\\=\frac{sin\alpha(cos^2\alpha-sin^2\alpha)}{cos\alpha(cos^2\alpha-sin^2\alpha)}=\frac{sin\alpha}{cos\alpha}=tan\alpha\\\\\\sin^2\alpha+cos^2\alpha=1\to cos^2\alpha=1-sin^2\alpha\\sin^2\alpha+cos^2\alpha=1\to sin^2\alpha=1-cos^2\alpha\to-sin^2\alpha=cos^2\alpha-1$