Answer to Question #88929 in Trigonometry for Manu Chemjong

Question

cos^3a sin3a+sin^3a cos3a=3/4sin4a

Answer 1

$$\text{sin}^3\alpha \text{cos}(3\alpha) + \text{cos}^3\alpha \text{sin}(3\alpha)=$$

$$=\text{sin}^3\alpha(4\text{cos}^3\alpha-3\text{cos}\alpha)+\text{cos}^3\alpha(3\text{sin}\alpha-4\text{sin}^3\alpha)=$$

$$=4\text{sin}^3\alpha\text{cos}^3\alpha-3\text{sin}^3\alpha\text{cos}\alpha+3\text{cos}^3\alpha\text{sin}\alpha-4\text{cos}^3\alpha\text{sin}^3\alpha=$$

$$=3\text{sin}\alpha\text{cos}\alpha(\text{cos}^2\alpha-\text{sin}^2\alpha)=$$

$$=\frac{3}{2}\text{sin}(2\alpha)\text{cos}(2\alpha)=\frac{3}{4}\text{sin}(4\alpha)$$