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Question Number 42036 by Tawa1 last updated on 17/Aug/18
$$\mathrm{State}\:\mathrm{the}\:\mathrm{phase}\:\mathrm{shift},\:\:\mathrm{the}\:\mathrm{amplitude}\:\mathrm{and}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{graph}. \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{g}\left(\theta\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}\left(\mathrm{2}\theta\:+\:\pi\right) \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{f}\left(\theta\right)\:=\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}\left(\mathrm{2}\theta\:+\:\pi\right) \\ $$$$\left(\mathrm{c}\right)\:\:\mathrm{f}\left(\theta\right)\:=\:\mathrm{4}\theta \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 17/Aug/18
![a) g(θ) =−(3/4)sin(2θ) we have g(θ+π) =−(3/4)sin(2(θ+π)) =−(3/4)sin(2θ +2π) =−(3/4)sin(2θ) =g(θ) ⇒ g periodic with T=π we can study the variation on [−(π/2),(π/2)] but g(−θ) =(3/4)sin(2θ)=−g(θ) ⇒ g is odd so we can study the variation on [0,(π/2)] =[0,(π/4)]∪[(π/4),(π/2)] g^′ (θ) =−(3/2)cos(2θ) we have 2θ ∈[0,π] x 0 (π/4) (π/2) we have g(o)=0 , g((π/4))=−(3/4) g^′ (x) − + g((π/2)) =0 .... g(x) decr inc](Q42047.png)
$$\left.{a}\right)\:{g}\left(\theta\right)\:=−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)\:\:\:\:{we}\:{have}\:{g}\left(\theta+\pi\right)\:=−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{2}\left(\theta+\pi\right)\right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\:+\mathrm{2}\pi\right)\:=−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)\:={g}\left(\theta\right)\:\Rightarrow\:{g}\:{periodic}\:{with}\:{T}=\pi\: \\ $$$${we}\:{can}\:{study}\:{the}\:{variation}\:{on}\:\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\:\:\:{but}\:\:{g}\left(−\theta\right)\:=\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)=−{g}\left(\theta\right)\:\Rightarrow \\ $$$${g}\:{is}\:{odd}\:\:{so}\:{we}\:{can}\:{study}\:{the}\:{variation}\:{on}\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\:=\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\cup\left[\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}}\right] \\ $$$${g}^{'} \left(\theta\right)\:=−\frac{\mathrm{3}}{\mathrm{2}}{cos}\left(\mathrm{2}\theta\right)\:\:\:\:{we}\:{have}\:\:\mathrm{2}\theta\:\in\left[\mathrm{0},\pi\right]\: \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{we}\:{have}\:{g}\left({o}\right)=\mathrm{0}\:\:\:\:\:,\:{g}\left(\frac{\pi}{\mathrm{4}}\right)=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${g}^{'} \left({x}\right)\:\:\:\:\:\:\:\:−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left(\frac{\pi}{\mathrm{2}}\right)\:=\mathrm{0}\:\:\:.... \\ $$$${g}\left({x}\right)\:\:\:\:\:\:\:\:{decr}\:\:\:\:\:\:\:\:\:\:{inc} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
$${steps}\:{to}\:{draw}... \\ $$$$\left.\mathrm{1}\right){draw}\:{sin}\theta\:\:{in}\:{between}\:\:\theta=\:\mathrm{0}\:{and}\:\mathrm{2}\Pi...{we}\:{know} \\ $$$${the}\:{shape}\:{of}\:{sin}\theta\:{then}\:{repeate}\:{the}\:{shape}\:{but} \\ $$$$−{ve}\:{side}.. \\ $$$$\left.\mathrm{2}\right){f}\left(\theta\right)\:\mathrm{0}\:\:\:\mathrm{30}^{{o}} \:\:\mathrm{45}^{{o}} \:\:\mathrm{60}^{{o}} \:\:\mathrm{90}^{{o}} \:\:\mathrm{120}^{{o}} \:\:\mathrm{135}^{{o}} \:\:\mathrm{150}^{{o}} \:\mathrm{180}^{{o}} \\ $$$$\:\:\:\:\:{sin}\theta\:\mathrm{0}\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\:\:\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\:\:\:\mathrm{1}\:\:\:\:\:\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\mathrm{0} \\ $$$${now}\:{i}\:{attaching}\:\:{to}\:{clarify}... \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
$${if}\:{you}\:{wantfree}\:{books}\:{pdf}\:{form}\:{download}\:{able} \\ $$$${seach}\:{archives}.{org}\:{then}\:{in}\:{search}\:{option}\:{write} \\ $$$${physics}\:{or}\:{mathematics}\:{or}\:{graph}\:{or}\:{algebra} \\ $$$${here}\:{you}\:{get}\:{irodov}\:{krotov}\:{sl}\:{loneh}\: \\ $$$${algebra}\:{hall}\:{and}\:{knight}\:{etc}... \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{will}\:\mathrm{try}\:\mathrm{it}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\: \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{It}\:\mathrm{is}\:\mathrm{good}\:\mathrm{sir}.\:\mathrm{thanks}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$