sin θ=sin αsin (((θ+α)/2)) Express θ explicitly in terms of α.


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Question Number 39486 by ajfour last updated on 06/Jul/18

$\sin θ = \sin α \sin (\frac{θ + α}{2})$ $E x p r e s s θ e x p l i c i t l y i n t e r m s o f α .$
Commented by math khazana by abdo last updated on 07/Jul/18
$⇒2 sin((θ/2))cos((θ/2))=sinα(sin((θ/2))cos((α/2))+ cos((θ/2))sin((α/2))) let put sin((θ/2))=x ⇒ 2xξ(√(1−x^2 )) = sinα( x cos((α/2)) +ξ(√(1−x^2 ))sin((α/2)))⇒ 2xξ(√(1−x^2 ))=sinα cos((α/2))x +sinαsin((α/2))ξ(√(1−x^2 )) (2xξ−sin(α)sin((α/2))ξ)(√(1−x^2 ))=sin(α)cos((α/2))x⇒ (2x −sin(α)sin((α/2)))^2 (1−x^2 )=sin^2 (α)cos^2 ((α/2))x^2 4x^2 −4sin(α)sin((α/2))x +sin^2 (α)sin^2 ((α/2)) −sin^2 (α)cos^2 ((α/2))x^2 =0 ⇒ (4−sin^2 (α)cos^2 ((α/2)))x^2 −4 sin(α)sin((α/2))x +sin^2 (α)sin^2 ((α/2)) =0 Δ^′ =4 sin^2 (α)sin^2 ((α/2))−(4−sin^2 (α)cos^2 ((α/2)))(sin^2 αsin^2 ((α/2))) =4 sin^2 (α)sin^2 ((α/2)) −4sin^2 (α)sin^2 ((α/2)) +sin^4 (α) sin^2 ((α/2))cos^2 ((α/2)) =(1/4) sin^6 (α) ⇒x=((2sin(α)sin((α/2))+^− (1/2)∣sinα∣^3 )/(4−sin^2 (α)cos^2 ((α/2)))) ⇒ θ =2arcsin(x)⇒ θ =2arcsin{((2sin(α)sin((α/2)) +^− (1/2)∣sinα∣^3 )/(4−sin^2 (α)cos^2 ((α/2))))}.$
$\Rightarrow 2 s i n (\frac{θ}{2}) c o s (\frac{θ}{2}) = s i n α (s i n (\frac{θ}{2}) c o s (\frac{α}{2}) +$ $c o s (\frac{θ}{2}) s i n (\frac{α}{2})) l e t p u t s i n (\frac{θ}{2}) = x \Rightarrow$ $2 x ξ \sqrt{1 - x^{2}} = s i n α (x c o s (\frac{α}{2}) + ξ \sqrt{1 - x^{2}} s i n (\frac{α}{2})) \Rightarrow$ $2 x ξ \sqrt{1 - x^{2}} = s i n α c o s (\frac{α}{2}) x + s i n α s i n (\frac{α}{2}) ξ \sqrt{1 - x^{2}}$ $(2 x ξ - s i n (α) s i n (\frac{α}{2}) ξ) \sqrt{1 - x^{2}} = s i n (α) c o s (\frac{α}{2}) x \Rightarrow$ ${(2 x - s i n (α) s i n (\frac{α}{2}))}^{2} (1 - x^{2}) = {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2}) x^{2}$ $4 x^{2} - 4 s i n (α) s i n (\frac{α}{2}) x + {s i n}^{2} (α) {s i n}^{2} (\frac{α}{2})$ $- {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2}) x^{2} = 0 \Rightarrow$ $(4 - {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2})) x^{2} - 4 s i n (α) s i n (\frac{α}{2}) x$ $+ {s i n}^{2} (α) {s i n}^{2} (\frac{α}{2}) = 0$ $Δ^{^{'}} = 4 {s i n}^{2} (α) {s i n}^{2} (\frac{α}{2}) - (4 - {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2})) ({s i n}^{2} α {s i n}^{2} (\frac{α}{2}))$ $= 4 {s i n}^{2} (α) {s i n}^{2} (\frac{α}{2}) - 4 {s i n}^{2} (α) {s i n}^{2} (\frac{α}{2})$ $+ {s i n}^{4} (α) {s i n}^{2} (\frac{α}{2}) {c o s}^{2} (\frac{α}{2})$ $= \frac{1}{4} {s i n}^{6} (α) \Rightarrow x = \frac{2 s i n (α) s i n (\frac{α}{2}) \bar{+} \frac{1}{2} ∣ s i n α ∣^{3}}{4 - {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2})} \Rightarrow$ $θ = 2 a r c s i n (x) \Rightarrow$ $θ = 2 a r c s i n {\frac{2 s i n (α) s i n (\frac{α}{2}) \bar{+} \frac{1}{2} ∣ s i n α ∣^{3}}{4 - {s i n}^{2} (α) {c o s}^{2} (\frac{α}{2})}} .$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

$e x c e l l e n t q u e s t i o n . . p l s g i v e t i m e . . .$
Commented by math khazana by abdo last updated on 07/Jul/18

$ξ^{2} = 1$
Commented by ajfour last updated on 07/Jul/18

Commented by ajfour last updated on 07/Jul/18

$I f w e t a k e ∠ C A D t o b e θ a n d$ $e x p r e s s θ i n t e r m s o f α . W e t h e n$ $o b t a i n a i n t e r m s o f r a d i u s R$ $a n d α (w h i c h w a s t h e q u e s t i o n) .$
Commented by ajfour last updated on 07/Jul/18

$T h a n k y o u S i r, l e t m e s e e i f i c a n$ $f o l l o w y o u r s o l u t i o n .$
Commented by ajfour last updated on 08/Jul/18

$l i n e 6 t o l i n e 7 h o w S i r, p l e a s e c h e c k;$ $w h a t a b o u t t e r m w i t h x^{4} a n d x^{3} ?$
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