Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 42036 by Tawa1 last updated on 17/Aug/18

$$\mathrm{State}\mathrm{the}\mathrm{phase}\mathrm{shift},\mathrm{the}\mathrm{amplitude}\mathrm{and}\mathrm{draw}\mathrm{the}\mathrm{graph}.$$$$\left(\mathrm{a}\right)\mathrm{g}\left(\theta \right)=\frac{3}{4}\sin (2\theta +\pi )$$$$\left(\mathrm{b}\right)\mathrm{f}\left(\theta \right)=1+\frac{3}{4}\sin (2\theta +\pi )$$$$\left(\mathrm{c}\right)\mathrm{f}\left(\theta \right)=4\theta$$$$$$

Commented by maxmathsup by imad last updated on 17/Aug/18

$$a)g\left(\theta \right)=-\frac{3}{4}sin\left(2\theta \right)wehaveg(\theta +\pi )=-\frac{3}{4}sin\left(2(\theta +\pi )\right)$$$$=-\frac{3}{4}sin(2\theta +2\pi )=-\frac{3}{4}sin\left(2\theta \right)=g\left(\theta \right)\Rightarrow gperiodicwithT=\pi$$$$wecanstudythevariationon[-\frac{\pi }{2},\frac{\pi }{2}]butg(-\theta )=\frac{3}{4}sin\left(2\theta \right)=-g\left(\theta \right)\Rightarrow$$$$gisoddsowecanstudythevariationon[0,\frac{\pi }{2}]=[0,\frac{\pi }{4}]\cup [\frac{\pi }{4},\frac{\pi }{2}]$$$$g^{{}^{\prime }}\left(\theta \right)=-\frac{3}{2}cos\left(2\theta \right)wehave2\theta \in [0,\pi ]$$$$x0\frac{\pi }{4}\frac{\pi }{2}wehaveg\left(o\right)=0,g\left(\frac{\pi }{4}\right)=-\frac{3}{4}$$$$g^{{}^{\prime }}\left(x\right)-+g\left(\frac{\pi }{2}\right)=0....$$$$g\left(x\right)decrinc$$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

$$stepstodraw...$$$$1)drawsin\theta inbetween\theta =0and2\mathrm{\Pi }...weknow$$$$theshapeofsin\theta thenrepeatetheshapebut$$$$-veside..$$$$2)f\left(\theta \right)0{30}^o{45}^o{60}^o{90}^o{120}^o{135}^o{150}^o{180}^o$$$$sin\theta 0\frac{1}{2}\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}1\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}\frac{1}{2}0$$$$nowiattachingtoclarify...$$

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

$$ifyouwantfreebookspdfformdownloadable$$$$seacharchives.orgtheninsearchoptionwrite$$$$physicsormathematicsorgraphoralgebra$$$$hereyougetirodovkrotovslloneh$$$$algebrahallandknightetc...$$

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

$$\mathrm{It}\mathrm{is}\mathrm{good}\mathrm{sir}.\mathrm{thanks}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com