Menu Close

Prove-that-2sin-2-cos-2-sin-sin-I-need-help-immediately-please-

Tinku Tara 0 Comments
Pin




Question Number 126562 by mey3nipaba last updated on 21/Dec/20
Prove that 2sin ((θ+φ)/2)cos ((θ−φ)/2)=sin θ+sin ∅ I need help immediately please
$${Prove}\:{that}\:\mathrm{2sin}\:\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\:\frac{\theta−\phi}{\mathrm{2}}=\mathrm{sin}\:\theta+\mathrm{sin}\:\emptyset \\ $$$${I}\:{need}\:{help}\:{immediately}\:{please} \\ $$
Commented by mr W last updated on 21/Dec/20
θ=((θ+ϕ)/2)+((θ−ϕ)/2) ϕ=((θ+ϕ)/2)−((θ−ϕ)/2) sin θ=sin (((θ+ϕ)/2)+((θ−ϕ)/2))=sin ((θ+ϕ)/2) cos ((θ−ϕ)/2)+cos ((θ+ϕ)/2) sin ((θ−ϕ)/2) sin ϕ=sin (((θ+ϕ)/2)−((θ−ϕ)/2))=sin ((θ+ϕ)/2) cos ((θ−ϕ)/2)−cos ((θ+ϕ)/2) sin ((θ−ϕ)/2) ⇒sin θ+sin ϕ=2 sin ((θ+ϕ)/2) cos ((θ−ϕ)/2)
$$\theta=\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\varphi=\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta=\mathrm{sin}\:\left(\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}}\right)=\mathrm{sin}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}}+\mathrm{cos}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{sin}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\varphi=\mathrm{sin}\:\left(\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}}\right)=\mathrm{sin}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}}−\mathrm{cos}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{sin}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi=\mathrm{2}\:\mathrm{sin}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$
Answered by Olaf last updated on 21/Dec/20
Many ways to prove that. Example : Let f_φ (θ) = 2sin((θ+φ)/2)cos((θ−φ)/2) and g_φ (θ) = sinθ+sinφ (φ is a parameter) f_φ ′(θ) = 2[(1/2)cos((θ+φ)/2)cos((θ−φ)/2)−(1/2)sin((θ+φ)/2)cos((θ+φ)/2)] f_φ ′(θ) = cos(((θ+φ)/2)+((θ−φ)/2)) = cosθ = g_φ ′(θ) ⇒ f_φ (θ) = g_φ (θ)+C C = f_φ (φ) −g_φ (φ) = 2sinφ−2sinφ = 0 f_φ (θ) = g_φ (θ)
$$\mathrm{Many}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}. \\ $$$$\mathrm{Example}\:: \\ $$$$\mathrm{Let}\:{f}_{\phi} \left(\theta\right)\:=\:\mathrm{2sin}\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\frac{\theta−\phi}{\mathrm{2}} \\ $$$$\mathrm{and}\:{g}_{\phi} \left(\theta\right)\:=\:\mathrm{sin}\theta+\mathrm{sin}\phi \\ $$$$\left(\phi\:\mathrm{is}\:\mathrm{a}\:\mathrm{parameter}\right) \\ $$$${f}_{\phi} '\left(\theta\right)\:=\:\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\frac{\theta−\phi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\frac{\theta+\phi}{\mathrm{2}}\mathrm{cos}\frac{\theta+\phi}{\mathrm{2}}\right] \\ $$$${f}_{\phi} '\left(\theta\right)\:=\:\mathrm{cos}\left(\frac{\theta+\phi}{\mathrm{2}}+\frac{\theta−\phi}{\mathrm{2}}\right)\:=\:\mathrm{cos}\theta\:=\:{g}_{\phi} '\left(\theta\right) \\ $$$$\Rightarrow\:{f}_{\phi} \left(\theta\right)\:=\:{g}_{\phi} \left(\theta\right)+\mathrm{C} \\ $$$$\mathrm{C}\:=\:{f}_{\phi} \left(\phi\right)\:−{g}_{\phi} \left(\phi\right)\:=\:\mathrm{2sin}\phi−\mathrm{2sin}\phi\:=\:\mathrm{0} \\ $$$${f}_{\phi} \left(\theta\right)\:=\:{g}_{\phi} \left(\theta\right) \\ $$
Answered by physicstutes last updated on 21/Dec/20
lets begin with sin (A + B) = sin A cos B + sin B cos A......(x) sin (A−B) = sin A cos B − sin B cos A.......(y) let A + B = θ .....(i) and A−B = ∅.....(ii) (ii) + (i) ⇒ A = ((θ + ∅)/2) similarly B = ((θ−∅)/2) (x) + (y) ⇒ sin ∅ + sin θ = 2 sin A cos B ⇒ sin θ + sin ∅ = 2 sin (((θ + ∅)/2)) cos (((θ−∅)/2))
$$\mathrm{lets}\:\mathrm{begin}\:\mathrm{with} \\ $$$$\:\mathrm{sin}\:\left({A}\:+\:{B}\right)\:=\:\mathrm{sin}\:{A}\:\mathrm{cos}\:{B}\:+\:\mathrm{sin}\:{B}\:\mathrm{cos}\:{A}……\left(\mathrm{x}\right) \\ $$$$\:\mathrm{sin}\:\left({A}−{B}\right)\:=\:\mathrm{sin}\:{A}\:\mathrm{cos}\:{B}\:−\:\mathrm{sin}\:{B}\:\mathrm{cos}\:{A}…….\left(\mathrm{y}\right) \\ $$$$\mathrm{let}\:{A}\:+\:{B}\:=\:\theta\:…..\left({i}\right) \\ $$$$\:\mathrm{and}\:{A}−{B}\:=\:\emptyset…..\left({ii}\right) \\ $$$$\:\left({ii}\right)\:+\:\left({i}\right)\:\Rightarrow\:{A}\:=\:\frac{\theta\:+\:\emptyset}{\mathrm{2}}\:\mathrm{similarly}\:\:{B}\:=\:\frac{\theta−\emptyset}{\mathrm{2}} \\ $$$$\:\left(\mathrm{x}\right)\:+\:\left(\mathrm{y}\right)\:\Rightarrow\:\:\mathrm{sin}\:\emptyset\:+\:\mathrm{sin}\:\theta\:=\:\mathrm{2}\:\mathrm{sin}\:{A}\:\mathrm{cos}\:{B} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:\theta\:+\:\mathrm{sin}\:\emptyset\:=\:\mathrm{2}\:\mathrm{sin}\:\left(\frac{\theta\:+\:\emptyset}{\mathrm{2}}\right)\:\mathrm{cos}\:\left(\frac{\theta−\emptyset}{\mathrm{2}}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *