By writing sin3A \sin 3A\,sin3A as sin(2A+A)\sin(2A + A)sin(2A+A), show that sin3A=3sinA−4sin3A\sin 3A = 3\sin A - 4\sin^3 Asin3A=3sinA−4sin3A
Solve, for 0≤A≤π0 \leq A \leq \pi0≤A≤π, the equation,
3sinA−4sin3A=12 3\sin A - 4\sin^3 A = \frac{1}{\sqrt{2}} 3sinA−4sin3A=21Give your answers in terms of π\piπ.