$\cos x +i(1+\sin x)$

I already know that: $\bbox[yellow] {\sin x=2\sin(\frac{x}{2})\cos(\frac{x}{2})}$ and $\bbox[yellow]{\cos x=2\cos^2(\frac{x}{2})-1}$

But this doesn't help me in this case.


Using your tips and $\cos(\pi/2-x)=\sin(x)$,

$$\cos x+i(1+\sin x)=\sin(\frac\pi2-x)+i(1+\cos(\frac\pi2-x))\\ =2\sin(\frac\pi4-\frac x2)\cos(\frac\pi4-\frac x2)+2i\cos^2(\frac\pi4-\frac x2)\\ =2\cos(\frac\pi4-\frac x2)\left[\sin(\frac\pi4-\frac x2)+i\cos(\frac\pi4-\frac x2)\right]\\ =2\cos(\frac\pi4-\frac x2)\left[\cos(\frac\pi4+\frac x2)+i\sin(\frac\pi4+\frac x2)\right]\\ =2\cos(\frac\pi 4-\frac x2)\exp(i(\frac\pi4+\frac x2))$$

  • $\begingroup$ Thank you very much. Your answer is clear and helpful. Have a nice day. $\endgroup$ – Zakaria Feb 15 at 14:35

Let $$z=\cos x+(1+\sin x)i=\rho \,e^{i\phi}.$$ You can use the identity $e^{i\phi}=\cos\phi+i\sin\phi$ and demand that both hand sides of the first relation match. That is, the real part of the RHS should match with the real part of the LHS and the same for the imaginary parts. Upon doing so, you will find a $2\times2$ system of algebraic equations involving $\rho$ and $\phi$. Then you can solve it and obtain what you want.


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