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Question Number 39486 by ajfour last updated on 06/Jul/18
sin θ=sin αsin (((θ+α)/2)) Express θ explicitly in terms of α.
$$\mathrm{sin}\:\theta=\mathrm{sin}\:\alpha\mathrm{sin}\:\left(\frac{\theta+\alpha}{\mathrm{2}}\right) \\ $$$${Express}\:\theta\:{explicitly}\:{in}\:{terms}\:{of}\:\alpha. \\ $$
Commented by math khazana by abdo last updated on 07/Jul/18
⇒2 sin((θ/2))cos((θ/2))=sinα(sin((θ/2))cos((α/2))+ cos((θ/2))sin((α/2))) let put sin((θ/2))=x ⇒ 2xξ(√(1−x^2 )) = sinα( x cos((α/2)) +ξ(√(1−x^2 ))sin((α/2)))⇒ 2xξ(√(1−x^2 ))=sinα cos((α/2))x +sinαsin((α/2))ξ(√(1−x^2 )) (2xξ−sin(α)sin((α/2))ξ)(√(1−x^2 ))=sin(α)cos((α/2))x⇒ (2x −sin(α)sin((α/2)))^2 (1−x^2 )=sin^2 (α)cos^2 ((α/2))x^2 4x^2 −4sin(α)sin((α/2))x +sin^2 (α)sin^2 ((α/2)) −sin^2 (α)cos^2 ((α/2))x^2 =0 ⇒ (4−sin^2 (α)cos^2 ((α/2)))x^2 −4 sin(α)sin((α/2))x +sin^2 (α)sin^2 ((α/2)) =0 Δ^′ =4 sin^2 (α)sin^2 ((α/2))−(4−sin^2 (α)cos^2 ((α/2)))(sin^2 αsin^2 ((α/2))) =4 sin^2 (α)sin^2 ((α/2)) −4sin^2 (α)sin^2 ((α/2)) +sin^4 (α) sin^2 ((α/2))cos^2 ((α/2)) =(1/4) sin^6 (α) ⇒x=((2sin(α)sin((α/2))+^− (1/2)∣sinα∣^3 )/(4−sin^2 (α)cos^2 ((α/2)))) ⇒ θ =2arcsin(x)⇒ θ =2arcsin{((2sin(α)sin((α/2)) +^− (1/2)∣sinα∣^3 )/(4−sin^2 (α)cos^2 ((α/2))))}.
$$\Rightarrow\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right)={sin}\alpha\left({sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)+\right. \\ $$$$\left.{cos}\left(\frac{\theta}{\mathrm{2}}\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right)\right)\:\:{let}\:{put}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)={x}\:\Rightarrow \\ $$$$\mathrm{2}{x}\xi\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:{sin}\alpha\left(\:{x}\:{cos}\left(\frac{\alpha}{\mathrm{2}}\right)\:+\xi\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\right)\Rightarrow \\ $$$$\mathrm{2}{x}\xi\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={sin}\alpha\:{cos}\left(\frac{\alpha}{\mathrm{2}}\right){x}\:+{sin}\alpha{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\xi\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}{x}\xi−{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right)\xi\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={sin}\left(\alpha\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right){x}\Rightarrow \\ $$$$\left(\mathrm{2}{x}\:−{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right)\right)^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)={sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right){x}^{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{4}{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right){x}\:+{sin}^{\mathrm{2}} \left(\alpha\right){sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$−{sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right){x}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{4}−{sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\right){x}^{\mathrm{2}} \:−\mathrm{4}\:{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right){x} \\ $$$$+{sin}^{\mathrm{2}} \left(\alpha\right){sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\:=\mathrm{0} \\ $$$$\Delta^{'} \:\:=\mathrm{4}\:{sin}^{\mathrm{2}} \left(\alpha\right){sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)−\left(\mathrm{4}−{sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\right)\left({sin}^{\mathrm{2}} \alpha{sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{4}\:{sin}^{\mathrm{2}} \left(\alpha\right){sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\:−\mathrm{4}{sin}^{\mathrm{2}} \left(\alpha\right){sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\: \\ $$$$+{sin}^{\mathrm{4}} \left(\alpha\right)\:{sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:{sin}^{\mathrm{6}} \left(\alpha\right)\:\Rightarrow{x}=\frac{\mathrm{2}{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right)\overset{−} {+}\frac{\mathrm{1}}{\mathrm{2}}\mid{sin}\alpha\mid^{\mathrm{3}} }{\mathrm{4}−{sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\theta\:=\mathrm{2}{arcsin}\left({x}\right)\Rightarrow \\ $$$$\theta\:=\mathrm{2}{arcsin}\left\{\frac{\mathrm{2}{sin}\left(\alpha\right){sin}\left(\frac{\alpha}{\mathrm{2}}\right)\:\overset{−} {+}\:\frac{\mathrm{1}}{\mathrm{2}}\mid{sin}\alpha\mid^{\mathrm{3}} }{\mathrm{4}−{sin}^{\mathrm{2}} \left(\alpha\right){cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)}\right\}. \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
excellent question..pls give time ...
$${excellent}\:{question}..{pls}\:{give}\:{time}\:… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 07/Jul/18
ξ^2 =1
$$\xi^{\mathrm{2}} \:=\mathrm{1} \\ $$
Commented by ajfour last updated on 07/Jul/18
Commented by ajfour last updated on 07/Jul/18
If we take ∠CAD to be θ and express θ in terms of α. We then obtain a in terms of radius R and 𝛂 (which was the question).
$${If}\:{we}\:{take}\:\angle{CAD}\:{to}\:{be}\:\theta\:{and} \\ $$$${express}\:\:\theta\:{in}\:{terms}\:{of}\:\alpha.\:{We}\:{then} \\ $$$${obtain}\:\boldsymbol{{a}}\:{in}\:{terms}\:{of}\:{radius}\:{R} \\ $$$${and}\:\boldsymbol{\alpha}\:\left({which}\:{was}\:{the}\:{question}\right). \\ $$
Commented by ajfour last updated on 07/Jul/18
Thank you Sir, let me see if i can follow your solution.
$${Thank}\:{you}\:{Sir},\:{let}\:{me}\:{see}\:{if}\:{i}\:{can} \\ $$$${follow}\:{your}\:{solution}. \\ $$
Commented by ajfour last updated on 08/Jul/18
line 6 to line 7 how Sir, please check; what about term with x^4 and x^3 ?
$${line}\:\mathrm{6}\:{to}\:{line}\:\mathrm{7}\:{how}\:{Sir},\:{please}\:{check}; \\ $$$${what}\:{about}\:{term}\:{with}\:{x}^{\mathrm{4}} \:{and}\:{x}^{\mathrm{3}} \:\:? \\ $$

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