Question

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

Answer 1

`(sin\alpha-2sin^3\alpha)/(2cos^3\alpha-cos\alpha)=(sin\alpha(1-2sin^2\alpha))/(cos\alpha(2cos^2\alpha-1))=(sin\alpha(1-sin^2\alpha-sin^2\alpha))/(cos\alpha(cos^2\alpha+cos^2\alpha-1))\\\\=(sin\alpha(cos^2\alpha-sin^2\alpha))/(cos\alpha(cos^2\alpha-sin^2\alpha))=(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`