Proving the trigonometrical identities

Question

please prove this answer, step by step.. $$\cos A - \cos 3A = 4 \sin^2A \cos A$$ I had just finished the left side $= -2 \sin 2A \sin A$ but then I have no idea to prove it..

Answer 1

$$\cos(A)-\cos(3A)=\cos(A)-(\cos(2A)\cos(A)-\sin(2A)\sin(A))$$

$$=\underbrace{\cos(A)-\cos^{3}(A)}_{=\cos(A)(1-\cos^{2}(A))}+\cos(A)\sin^{2}(A)+2\sin^{2}(A)\cos(A)$$

$$=\cos(A)\sin^{2}(A)+3\sin^{2}(A)\cos(A)=4\sin^{2}(A)\cos(A)$$

Answer 2

We use the identity $$\cos p-\cos q=-2\sin\left(\frac{p+q}{2}\right)\sin\left(\frac{p-q}{2}\right)$$ so with the double angle identity we have $$\cos A-\cos3A=\color{red}+2\sin(2A)\sin A=2\times 2\sin A\cos A\times \sin A=4\sin^2 A\cos A$$

Answer 3

$$\cos3A=4\cos^3A-3\cos A\implies\cos A-\cos3A=4\cos A(1-\cos^2A)$$

Answer 4

(for $A \in \mathbb{R})$ $$ \cos A - \cos 3A = \mathfrak{Re} (e^{iA} - e^{3iA}) \\ = \mathfrak{Re} \left(-2ie^{2iA}\frac{(e^{iA}-e^{-iA})}{2i} \right) \\ = \mathfrak{Im} \left(2e^{2iA} \right) \sin A \\ =2 \sin 2A \sin A \\ =4 \sin^2A \cos A $$