Question-223595

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Question Number 223595 by ajfour last updated on 31/Jul/25

Commented by ajfour last updated on 31/Jul/25

$$\:\:\:\:\:\:\:\:\:\:{Given}\:\theta.\:{Find}\:\phi. \\ $$

Answered by mr W last updated on 31/Jul/25

Commented by ajfour last updated on 31/Jul/25

C(sin φ, cos φ) A[tan ((π/4)+(θ/2)),0] B[tan ((π/4)+(θ/2)), tan θ tan ((π/4)+(θ/2))] ((1−cos φ)/(sin φ))=((1−tan θ tan ((π/4)+(θ/2)))/(tan ((π/4)+(θ/2)))) φ=2tan^(−1) (((1−tan θ tan ((π/4)+(θ/2)))/(tan ((π/4)+(θ/2)))))

$${C}\left(\mathrm{sin}\:\phi,\:\mathrm{cos}\:\phi\right) \\ $$$${A}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right),\mathrm{0}\right] \\ $$$${B}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right),\:\mathrm{tan}\:\theta\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)\right] \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\phi}{\mathrm{sin}\:\phi}=\frac{\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\phi=\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{tan}\:\theta\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)}\right) \\ $$$$ \\ $$

Commented by mr W last updated on 01/Aug/25

A(a, 0) B(a, b) C(sin φ, cos φ) ((cos θ)/a)+((sin θ)/1)=1 ⇒a=((cos θ)/(1−sin θ)) (b/a)=tan θ ⇒b=((sin θ)/(1−sin θ)) eqn. of DB: y=1−(((1−b)x)/a) cos φ=1−((1/a)−(b/a))sin φ cos φ=1−(((1−sin θ)/(cos θ))−tan θ)sin φ ((1−2 sin θ)/(cos θ)) sin φ=1−cos φ ((1−2 sin θ)/(cos θ))=tan (φ/2) ⇒φ=2 tan^(−1) ((1−2 sin θ)/(cos θ)) such that φ≥0: 0≤θ≤(π/6) for θ=0: φ=(π/2) for θ=(π/6): φ=0

$${A}\left({a},\:\mathrm{0}\right) \\ $$$${B}\left({a},\:{b}\right) \\ $$$${C}\left(\mathrm{sin}\:\phi,\:\mathrm{cos}\:\phi\right) \\ $$$$\frac{\mathrm{cos}\:\theta}{{a}}+\frac{\mathrm{sin}\:\theta}{\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta} \\ $$$$\frac{{b}}{{a}}=\mathrm{tan}\:\theta \\ $$$$\Rightarrow{b}=\frac{\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta} \\ $$$${eqn}.\:{of}\:{DB}: \\ $$$${y}=\mathrm{1}−\frac{\left(\mathrm{1}−{b}\right){x}}{{a}} \\ $$$$\mathrm{cos}\:\phi=\mathrm{1}−\left(\frac{\mathrm{1}}{{a}}−\frac{{b}}{{a}}\right)\mathrm{sin}\:\phi \\ $$$$\mathrm{cos}\:\phi=\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}−\mathrm{tan}\:\theta\right)\mathrm{sin}\:\phi \\ $$$$\frac{\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:\mathrm{sin}\:\phi=\mathrm{1}−\mathrm{cos}\:\phi \\ $$$$\frac{\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\mathrm{tan}\:\frac{\phi}{\mathrm{2}} \\ $$$$\Rightarrow\phi=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$${such}\:{that}\:\phi\geqslant\mathrm{0}:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{6}} \\ $$$${for}\:\theta=\mathrm{0}:\:\:\phi=\frac{\pi}{\mathrm{2}} \\ $$$${for}\:\theta=\frac{\pi}{\mathrm{6}}:\:\phi=\mathrm{0} \\ $$

Commented by ajfour last updated on 31/Jul/25

Sir, thank you. I think, for θ=π/4 i get 𝛗=−2tan^(−1) (2−(√2)) you made it real compact.

$${Sir},\:{thank}\:{you}.\:{I}\:{think},\:{for}\:\theta=\pi/\mathrm{4} \\ $$$${i}\:{get}\:\boldsymbol{\phi}=−\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{2}−\sqrt{\mathrm{2}}\right) \\ $$$${you}\:{made}\:{it}\:{real}\:{compact}. \\ $$

Commented by Tawa11 last updated on 01/Aug/25

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{wait}. \\ $$

Commented by mr W last updated on 31/Jul/25

Commented by mr W last updated on 31/Jul/25

Commented by mr W last updated on 31/Jul/25

Commented by Tawa11 last updated on 31/Jul/25

Please help with Q223591 sirs, when you are chanced. Thanks sir.

$$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{Q223591}\:\mathrm{sirs}, \\ $$$$\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chanced}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 31/Jul/25

tomorrow, sure.

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