Let alpha, beta and gamma be such that 0

Question

(D) Given that for any $x \in \mathbb{R}$,$\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0$. This represents the sum of three vectors of unit length

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) Given that for any $x \in \mathbb{R}$,$\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0$. This represents the sum of three vectors of unit length in the complex plane,which must form an equilateral triangle. Thus,the angles must be in arithmetic progression with a common difference of $\frac{2\pi}{3}$. Let $\beta - \alpha = \frac{2\pi}{3}$ and $\gamma - \beta = \frac{2\pi}{3}$. Then $\gamma - \alpha = (\gamma - \beta) + (\beta - \alpha) = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3}$. However,since $0 Alternatively,using the property of the sum of cosines: $\cos(x+\alpha) + \cos(x+\beta) + \cos(x+\gamma) = 0 \implies \alpha, \beta, \gamma$ are angles of a triangle or satisfy specific symmetry. For the sum to be zero for all $x$,the vectors $e^{i\alpha}, e^{i\beta}, e^{i\gamma}$ must sum to zero. This implies $\beta - \alpha = \frac{2\pi}{3}$ and $\gamma - \beta = \frac{2\pi}{3}$ (or vice versa). Thus,$\gamma - \alpha = \frac{4\pi}{3}$. Then $	an(\gamma - \alpha) = 	an(\frac{4\pi}{3}) = 	an(\pi + \frac{\pi}{3}) = 	an(\frac{\pi}{3}) = \sqrt{3}$.

Let $\alpha, \beta$ and $\gamma$ be such that $0 < \alpha < \beta < \gamma < 2 \pi$. For any $x \in \mathbb{R}$,if $\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0$,then $\tan (\gamma-\alpha) = $

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