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Question Number 21272 by Tinkutara last updated on 18/Sep/17

Let α, β ∈ (−π, π) be such that cos(α − β) = 1 and cos(α + β) = (1/e). The number of pairs of α, β satisfying the above system of equation is

$$\mathrm{Let}\:\alpha,\:\beta\:\in\:\left(−\pi,\:\pi\right)\:\mathrm{be}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{cos}\left(\alpha\:−\:\beta\right)\:=\:\mathrm{1}\:\mathrm{and}\:\mathrm{cos}\left(\alpha\:+\:\beta\right)\:=\:\frac{\mathrm{1}}{{e}}. \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{of}\:\alpha,\:\beta\:\mathrm{satisfying} \\ $$$$\mathrm{the}\:\mathrm{above}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{is} \\ $$

Answered by mrW1 last updated on 18/Sep/17

for α, β∈(−π,π) α−β ∈(−2π,2π) α+β ∈(−2π,2π) cos (α−β)=1 ⇒α−β=0 cos (α+β)=(1/e) with θ=cos^(−1) (1/e) and 0<θ<(π/2) ⇒α+β=±θ, ±(2π−θ) ⇒there are 4 pairs of α and β: α=(θ/2), β=(θ/2) α=−(θ/2), β=−(θ/2) α=π−(θ/2), β=π−(θ/2) α=−π+(θ/2), β=−π+(θ/2)

$$\mathrm{for}\:\alpha,\:\beta\in\left(−\pi,\pi\right) \\ $$$$\alpha−\beta\:\in\left(−\mathrm{2}\pi,\mathrm{2}\pi\right) \\ $$$$\alpha+\beta\:\in\left(−\mathrm{2}\pi,\mathrm{2}\pi\right) \\ $$$$ \\ $$$$\mathrm{cos}\:\left(\alpha−\beta\right)=\mathrm{1} \\ $$$$\Rightarrow\alpha−\beta=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}}{\mathrm{e}} \\ $$$$\mathrm{with}\:\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{e}}\:\mathrm{and}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\alpha+\beta=\pm\theta,\:\pm\left(\mathrm{2}\pi−\theta\right) \\ $$$$ \\ $$$$\Rightarrow\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{pairs}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\beta: \\ $$$$\alpha=\frac{\theta}{\mathrm{2}},\:\beta=\frac{\theta}{\mathrm{2}} \\ $$$$\alpha=−\frac{\theta}{\mathrm{2}},\:\beta=−\frac{\theta}{\mathrm{2}} \\ $$$$\alpha=\pi−\frac{\theta}{\mathrm{2}},\:\beta=\pi−\frac{\theta}{\mathrm{2}} \\ $$$$\alpha=−\pi+\frac{\theta}{\mathrm{2}},\:\beta=−\pi+\frac{\theta}{\mathrm{2}} \\ $$

Commented by Tinkutara last updated on 18/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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