How to solve $\sin (2z) =2\sin z \cos z$ and $\cos (2z) =\cos^2 (z) − \sin^2 (z)$

Question

How to solve this question?

I have tried substituting $z=x+iy$ but I couldnt get the answer.

Answer 1

These identities are best thought of as consequences of the angle-sum identities. For example, we know that $$ \sin(A+B) = \sin(A)\cos(B) + \cos(A) \sin(B). $$ This formula can be proven either using the identities $$ \cos(z) = \frac{e^{iz} + e^{-iz}}{2}, \quad \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} $$ that others have referred to, or using the geometric proof indicated by the first image in the section linked above (though the fact that this identity can simply be extended to complex arguments does require some justification). Once that first formula is established, setting $A = B = z$ yields $$ \sin(2z) = \sin(z+z) = \sin(z)\cos(z) + \cos(z) \sin(z) = 2 \sin(z) \cos(z). $$

Answer 2

Rearranging $e^{\pm iz}=\cos z\pm i\sin z$ gives $\cos z=\frac{e^{iz}+e^{-iz}}{2},\,\sin z=\frac{e^{iz}-e^{-iz}}{2i}$. So$$2\sin z\cos z=\frac{e^{2iz}-e^{-2iz}}{2i}=\sin 2z,\\\cos^2z-\sin^2z=\frac{(e^{iz}+e^{-iz})^2+(e^{iz}-e^{-iz})^2}{4}=\frac{e^{2iz}+e^{-2iz}}{2}=\cos 2z.$$