Given:
Given the steps of the proof of the equation
[tex]\sin\theta-\sin^3\theta=\frac{2\sin2\theta\cos^2\theta}{2\cos\theta}[/tex]Required: Expression missing on the thrd step
Explanation:
The second step is
[tex]\sin\theta-\sin^3\theta=\sin\theta(1-\sin^2\theta)\frac{2\cos\theta}{2\cos\theta}[/tex]from which leads to
[tex]\sin\theta-\sin^3\theta=\frac{(2\sin\theta\cos\theta)(1-\sin^2\theta)}{2\cos\theta}[/tex]The expression missing on the third step is
[tex]\frac{(2\sin\theta\cos\theta)(1-\sin^2\theta)}{2\cos\theta}[/tex]Option D is correct.
Final Answer:
[tex]\frac{(2\sin\theta\cos\theta)(1-\sin^2\theta)}{2\cos\theta}[/tex]