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If $\beta,\gamma\in(0,\pi)$ such that $\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+\beta+\gamma)=0$ and $\sin\alpha+\sin(\alpha+\beta)+\sin(\alpha+\beta+\gamma)=0$; and $f(x)=\frac{\sin2x}{1+\cos2x}$, $g(x)=\frac{1+\sin x-\cos x}{1+\sin x+\cos x}$, Find $f'(\beta)+\lim_{x\to\gamma}g(x)$

Attempt:

$$f(x)=\tan x$$ $$g(x)=\tan\frac{x}{2}$$ $$f'(x)=\sec^2x$$

How can I compute $\gamma$?

Aditya Dev
  • 4,774

1 Answers1

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Using de' Moivre's Theorem

$$e^{i\alpha}+e^{i(\alpha+\beta)}+e^{i(\alpha+\beta+\gamma)}=0$$

$|e^{i\alpha}|=1, e^{i\alpha}\ne0$

$$\implies1+e^{i\beta}(1+e^{i\gamma})=0\iff e^{i\gamma/2}+e^{-i\gamma/2}=-e^{-i(\gamma/2+\beta)}$$

$$\iff2\cos\dfrac\gamma2=e^{i(\pi-\gamma/2-\beta)}=\cos(\pi-\gamma/2-\beta)+i\sin(\pi-\gamma/2-\beta)$$

As $\cos\dfrac\gamma2,\sin(\pi-\gamma/2-\beta)$ are real,

$\sin(\pi-\gamma/2-\beta)=0\iff\pi-\gamma/2-\beta=-n\pi$ where $n$ is any integer

$\iff\gamma=2(n+1)\pi-2\beta$

Can you take it from here?