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Question Number 39529 by Rio Mike last updated on 07/Jul/18

Commented by Rio Mike last updated on 07/Jul/18

$$ineedhelpguys.cansomeonepleasd$$$$explaintomewhentouseeach$$$$oftheabovequadrantsandhow?$$$$intrigonometryicansolvetheequation$$$$but{won}^{\prime }tknowthequadranttouseandhow.$$$$forexample:solvefor\theta theequation$$$$2{sin}^2\theta =sin\theta$$$$2{sin}^2\theta -sin\theta =0$$$$sin\theta (2sin\theta -1)=0$$$$Eithersin\theta =0or\theta ={sin}^{-1}\frac{1}{2}$$$$whatdoidonow?$$

Commented by Rio Mike last updated on 07/Jul/18

$$thanksguys$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

$$sin\theta (2sin\theta -1)=0$$$$sin\theta =0so\theta =0$$$$2sin\theta -1=0$$$$sin\theta =\frac{1}{2}=sin\frac{\mathrm{\Pi }}{6}$$$$\theta =\frac{\mathrm{\Pi }}{6}$$$$nowthistwovalueliesinfirstquaadrant$$$$sin\theta =+vewhen\theta liesinsecondquadrant$$$$sosin0alsocanbewrittenas\mathit{s}\mathit{i}\mathit{n}(\mathbf{\Pi }-0)$$$$\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{i}\mathit{s}\mathit{s}\mathit{i}\mathit{n}\mathbf{\Pi }$$$$next\mathit{s}\mathit{i}\mathit{n}\frac{\mathrm{\Pi }}{6}alsocanbewrittenas$$$$\mathit{s}\mathit{i}\mathit{n}(\mathbf{\Pi }-\frac{\mathbf{\Pi }}{6})...\mathit{s}\mathit{i}\mathit{n}\frac{5\mathrm{\Pi }}{6}$$$$thereisformulaplswaitishallpost$$$$sin\theta =sin\alpha$$$$\theta =n\mathrm{\Pi }+{(-1)}^n\alpha$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18

Commented by MJS last updated on 07/Jul/18

$$\mathrm{in}\mathrm{geometric}\mathrm{examples}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{think}$$$$\mathrm{which}\mathrm{solution}\mathrm{fits}.\mathrm{i}.\mathrm{e}.\mathrm{if}\mathrm{you}\mathrm{calculate}\mathrm{an}$$$$\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{it}\mathrm{must}\mathrm{fit}\mathrm{into}\mathrm{this}\mathrm{triangle}$$$$(\alpha +\beta +\gamma =180°)$$$$\mathrm{in}‘‘\mathrm{free}\mathrm{from}\mathrm{any}{\mathrm{praxis}}^{″}\mathrm{examples}{\mathrm{you}}^{\prime }\mathrm{ll}$$$$\mathrm{have}\mathrm{to}\mathrm{deal}\mathrm{with}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{possible}$$$$\mathrm{solutions},\mathrm{same}\mathrm{as}\mathrm{in}\mathrm{equations}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{or}$$$$\mathrm{higher}\mathrm{degree}.$$$$$$$$\mathrm{it}\mathrm{might}\mathrm{help}\mathrm{to}\mathrm{think}\mathrm{of}\mathrm{the}\mathrm{coordinates}$$$$\mathrm{of}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$x^2+y^2=1\Rightarrow$$$$\Rightarrow {\cos }^2\alpha +{\sin }^2\alpha =1\Rightarrow$$$$\Rightarrow P=\left(\begin{array}{c}\cos \alpha \\ \sin \alpha \end{array}\right)$$$$0°<\alpha <90°\Rightarrow P\mathrm{in}{1}^{\mathrm{st}}\mathrm{quadrant}$$$$90°<\alpha <180°\Rightarrow P\mathrm{in}{2}^{\mathrm{nd}}\mathrm{quadrant}$$$$180°<\alpha <270°\Rightarrow P\mathrm{in}{3}^{\mathrm{rd}}\mathrm{quadrant}$$$$180°<\alpha <360°\Rightarrow P\mathrm{in}{4}^{\mathrm{th}}\mathrm{quadrant}$$$$\alpha =0°=360°\Rightarrow P\mathrm{on}{x}^{+}$$$$\alpha =90°\Rightarrow P\mathrm{on}{y}^{+}$$$$\alpha =180°\Rightarrow P\mathrm{on}{x}^{-}$$$$\alpha =270°\Rightarrow P\mathrm{on}{y}^{-}$$$$$$$$\mathrm{electronic}\mathrm{calculators}\mathrm{handle}\mathrm{angles}\mathrm{in}$$$$]-180°;180°]\mathrm{instead}\mathrm{of}[0°;360°[\mathrm{or}\mathrm{in}$$$$\mathrm{r}\mathrm{adiant}]-\pi ;\pi ]\mathrm{instead}\mathrm{of}[0;2\pi [$$$$\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{be}\mathrm{careful}\mathrm{with}\mathrm{arc}-\mathrm{functions}$$$$(\mathrm{always}\mathrm{remember}\mathrm{what}\mathrm{was}\mathrm{the}\mathrm{question},$$$$\mathrm{or}/\mathrm{and}\mathrm{try}\mathrm{the}\mathrm{other}\mathrm{possibilities}\mathrm{to}\mathrm{find}\mathrm{out}$$$$\mathrm{which}\mathrm{answer}\mathrm{fits}\mathrm{into}\mathrm{the}\mathrm{given}\mathrm{example})$$$$$$$$0°45°90°135°180°225°270°315°360°$$$$0°45°90°135°180°-135°-90°-45°0°$$$$0\frac{\pi }{4}\frac{\pi }{2}\frac{3\pi }{4}\pi \frac{5\pi }{4}\frac{3\pi }{2}\frac{7\pi }{4}2\pi$$$$0\frac{\pi }{4}\frac{\pi }{2}\frac{3\pi }{4}\pi -\frac{3\pi }{4}-\frac{\pi }{2}-\frac{\pi }{4}0$$

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