cos (alpha ,-,beta ) = 1 and cos (alpha | vedclass

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Question

Given that $\operatorname{cost}(\alpha-\beta)=1$ and $\cos (\alpha+\beta)=1 / \mathrm{e}$

where $\alpha, \beta \in[-\pi, \pi]$

Now $\cos (\alpha

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Given that $\operatorname{cost}(\alpha-\beta)=1$ and $\cos (\alpha+\beta)=1 / \mathrm{e}$

where $\alpha, \beta \in[-\pi, \pi]$

Now $\cos (\alpha-\beta)=1 \Rightarrow \alpha-\beta=0$ or $\alpha=\beta$

$	herefore $ $\cos (\alpha+\beta)=1 / e \Rightarrow \cos 2 \alpha=1 / e$

$\because 0<1 / \mathrm{e}<1$ and $2 \alpha \in[-2 \pi, 2 \pi]$

Therefore, there will be four values of $\alpha$ in

$[-2 \pi, 2 \pi]$ and correspondingly four values of $\beta$

Hence, there are four sets of $(\alpha, \beta)$.

$cos (\alpha \,-\,\beta ) = 1$ and $cos (\alpha +\beta ) = 1/e$ , where $\alpha , \beta \in [-\pi , \pi ]$ . Number of pairs of $(\alpha ,\beta )$ which satisfy both the equations is

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