1-if-sin-x-k-sin-find-tan-and-k-then-find-in-0-2pi-when-k-1-2-and-pi-2-if-x-sin-t-and-y-cos-2t-show-that-d-2-y-dx-2-4-0-

Question Number 86041 by M±th+et£s last updated on 26/Mar/20

1)if sin(θ−x)=k sin(θ+α) find tan(θ) and k then find θ in[0,2π] when k=(1/2) and α=π 2)if x=sin(t) and y=cos(2t) show that (d^2 y/dx^2 )+4=0

$$\left.\mathrm{1}\right){if}\: \\ $$$${sin}\left(\theta−{x}\right)={k}\:{sin}\left(\theta+\alpha\right) \\ $$$${find}\:{tan}\left(\theta\right)\:{and}\:{k} \\ $$$$ \\ $$$${then}\:{find}\:\theta\:{in}\left[\mathrm{0},\mathrm{2}\pi\right]\:\:{when}\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\alpha=\pi \\ $$$$ \\ $$$$\left.\mathrm{2}\right){if}\:{x}={sin}\left({t}\right)\:\:{and}\:\:{y}={cos}\left(\mathrm{2}{t}\right) \\ $$$${show}\:{that} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}=\mathrm{0} \\ $$

Answered by Rio Michael last updated on 26/Mar/20

2) x = sin t and y= cos (2t) ⇒ (dx/dt) = cos t and (dy/dt) = −2sin (2t) ⇒ (dy/dx) = (dy/dt) . (dt/dx) = ((−2sin(2t))/(cos t)) = ((−2(2sin t cos t))/(cos t)) = −4sin t (d^2 y/dx^2 ) = (d/dx)((dy/dx)) (dt/dx) = −4cos t× (1/(cos t)) = −4 ⇒ (d^2 y/(dx^2 )) = −4 hence (d^2 y/dx^2 ) + 4 = 0

$$\left.\mathrm{2}\right)\:\:{x}\:=\:\mathrm{sin}\:{t}\:\:\mathrm{and}\:\:{y}=\:\mathrm{cos}\:\left(\mathrm{2}{t}\right) \\ $$$$\:\Rightarrow\:\frac{{dx}}{{dt}}\:=\:\mathrm{cos}\:{t}\:\:\mathrm{and}\:\frac{{dy}}{{dt}}\:=\:−\mathrm{2sin}\:\left(\mathrm{2}{t}\right) \\ $$$$\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}\:.\:\frac{{dt}}{{dx}}\:=\:\frac{−\mathrm{2sin}\left(\mathrm{2}{t}\right)}{\mathrm{cos}\:{t}}\:=\:\frac{−\mathrm{2}\left(\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\right)}{\mathrm{cos}\:{t}}\:=\:−\mathrm{4sin}\:{t} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)\:\frac{{dt}}{{dx}}\:=\:−\mathrm{4cos}\:{t}×\:\frac{\mathrm{1}}{\mathrm{cos}\:{t}}\:=\:−\mathrm{4} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} \:\:}\:=\:−\mathrm{4}\:\:\mathrm{hence}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$

Commented by M±th+et£s last updated on 26/Mar/20