Question Number 39529 by Rio Mike last updated on 07/Jul/18
Commented by Rio Mike last updated on 07/Jul/18
$${i}\:{need}\:{help}\:{guys}.{can}\:{someone}\:{pleasd} \\ $$$${explain}\:{to}\:{me}\:\:{when}\:{to}\:{use}\:{each} \\ $$$${of}\:{the}\:{above}\:{quadrants}\:{and}\:{how}?\: \\ $$$${in}\:{trigonometry}\:{i}\:{can}\:{solve}\:{the}\:{equation} \\ $$$${but}\:{won}'{t}\:{know}\:{the}\:{quadrant}\:{to}\:{use}\:{and}\:{how}. \\ $$$$\:{for}\:{example}\::\:{solve}\:{for}\:\theta\:{the}\:{equation} \\ $$$$\:\mathrm{2}{sin}^{\mathrm{2}} \theta\:=\:{sin}\:\theta \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} \theta−\:{sin}\:\theta=\mathrm{0} \\ $$$${sin}\theta\left(\mathrm{2}{sin}\theta\:−\:\mathrm{1}\right)=\mathrm{0} \\ $$$${Either}\:{sin}\:\theta\:=\:\mathrm{0}\:{or}\:\theta\:=\:{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$${what}\:{do}\:{i}\:{do}\:{now}? \\ $$
Commented by Rio Mike last updated on 07/Jul/18
$${thanks}\:{guys} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
$${sin}\theta\left(\mathrm{2}{sin}\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$${sin}\theta=\mathrm{0}\:{so}\:\:\theta=\mathrm{0} \\ $$$$\mathrm{2}{sin}\theta−\mathrm{1}=\mathrm{0} \\ $$$${sin}\theta=\frac{\mathrm{1}}{\mathrm{2}}={sin}\frac{\Pi}{\mathrm{6}} \\ $$$$\theta=\frac{\Pi}{\mathrm{6}} \\ $$$${now}\:{this}\:{two}\:{value}\:{lies}\:{in}\:{first}\:{quaadrant} \\ $$$${sin}\theta=+{ve}\:{when}\:\theta\:{lies}\:{in}\:{second}\:{quadrant} \\ $$$${so}\:{sin}\mathrm{0}\:{also}\:{can}\:{be}\:{written}\:{as}\:\boldsymbol{{sin}}\left(\boldsymbol{\Pi}−\mathrm{0}\right) \\ $$$$\boldsymbol{{that}}\:\boldsymbol{{is}}\:\boldsymbol{{sin}\Pi} \\ $$$${next}\:\boldsymbol{{sin}}\frac{\Pi}{\mathrm{6}}\:{also}\:{can}\:{be}\:{written}\:{as}\: \\ $$$$\boldsymbol{{sin}}\left(\boldsymbol{\Pi}−\frac{\boldsymbol{\Pi}}{\mathrm{6}}\right)...\boldsymbol{{sin}}\frac{\mathrm{5}\Pi}{\mathrm{6}} \\ $$$${there}\:{is}\:{formula}\:{pls}\:{wait}\:{i}\:{shall}\:{post} \\ $$$${sin}\theta={sin}\alpha \\ $$$$\theta={n}\Pi+\left(−\mathrm{1}\right)^{{n}} \alpha \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
Commented by MJS last updated on 07/Jul/18
![in geometric examples you have to think which solution fits. i.e. if you calculate an angle of a triangle it must fit into this triangle (α+β+γ=180°) in “free from any praxis” examples you′ll have to deal with more than one possible solutions, same as in equations of 2^(nd) or higher degree. it might help to think of the coordinates of the points of the circle x^2 +y^2 =1 ⇒ ⇒ cos^2 α +sin^2 α=1 ⇒ ⇒ P= (((cos α)),((sin α)) ) 0°<α<90° ⇒ P in 1^(st) quadrant 90°<α<180° ⇒ P in 2^(nd) quadrant 180°<α<270° ⇒ P in 3^(rd) quadrant 180°<α<360° ⇒ P in 4^(th) quadrant α=0°=360° ⇒ P on x^+ α=90° ⇒ P on y^+ α=180° ⇒ P on x^− α=270° ⇒ P on y^− electronic calculators handle angles in ]−180°; 180°] instead of [0°; 360°[ or in radiant ]−π; π] instead of [0; 2π[ you have to be careful with arc−functions (always remember what was the question, or/and try the other possibilities to find out which answer fits into the given example) 0° 45° 90° 135° 180° 225° 270° 315° 360° 0° 45° 90° 135° 180° −135° −90° −45° 0° 0 (π/4) (π/2) ((3π)/4) π ((5π)/4) ((3π)/2) ((7π)/4) 2π 0 (π/4) (π/2) ((3π)/4) π −((3π)/4) −(π/2) −(π/4) 0](Q39542.png)
$$\mathrm{in}\:\mathrm{geometric}\:\mathrm{examples}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{think} \\ $$$$\mathrm{which}\:\mathrm{solution}\:\mathrm{fits}.\:\mathrm{i}.\mathrm{e}.\:\mathrm{if}\:\mathrm{you}\:\mathrm{calculate}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{it}\:\mathrm{must}\:\mathrm{fit}\:\mathrm{into}\:\mathrm{this}\:\mathrm{triangle} \\ $$$$\left(\alpha+\beta+\gamma=\mathrm{180}°\right) \\ $$$$\mathrm{in}\:``\mathrm{free}\:\mathrm{from}\:\mathrm{any}\:\mathrm{praxis}''\:\mathrm{examples}\:\mathrm{you}'\mathrm{ll} \\ $$$$\mathrm{have}\:\mathrm{to}\:\mathrm{deal}\:\mathrm{with}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one}\:\mathrm{possible} \\ $$$$\mathrm{solutions},\:\mathrm{same}\:\mathrm{as}\:\mathrm{in}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{or} \\ $$$$\mathrm{higher}\:\mathrm{degree}. \\ $$$$ \\ $$$$\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{to}\:\mathrm{think}\:\mathrm{of}\:\mathrm{the}\:\mathrm{coordinates} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\mathrm{sin}^{\mathrm{2}} \:\alpha=\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:{P}=\begin{pmatrix}{\mathrm{cos}\:\alpha}\\{\mathrm{sin}\:\alpha}\end{pmatrix} \\ $$$$\mathrm{0}°<\alpha<\mathrm{90}°\:\Rightarrow\:{P}\:\:\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{quadrant} \\ $$$$\mathrm{90}°<\alpha<\mathrm{180}°\:\Rightarrow\:{P}\:\:\mathrm{in}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{quadrant} \\ $$$$\mathrm{180}°<\alpha<\mathrm{270}°\:\Rightarrow\:{P}\:\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{quadrant} \\ $$$$\mathrm{180}°<\alpha<\mathrm{360}°\:\Rightarrow\:{P}\:\:\mathrm{in}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{quadrant} \\ $$$$\alpha=\mathrm{0}°=\mathrm{360}°\:\Rightarrow\:{P}\:\mathrm{on}\:{x}^{+} \\ $$$$\alpha=\mathrm{90}°\:\Rightarrow\:{P}\:\mathrm{on}\:{y}^{+} \\ $$$$\alpha=\mathrm{180}°\:\Rightarrow\:{P}\:\mathrm{on}\:{x}^{−} \\ $$$$\alpha=\mathrm{270}°\:\Rightarrow\:{P}\:\mathrm{on}\:{y}^{−} \\ $$$$ \\ $$$$\mathrm{electronic}\:\mathrm{calculators}\:\mathrm{handle}\:\mathrm{angles}\:\mathrm{in} \\ $$$$\left.\right]\left.−\mathrm{180}°;\:\mathrm{180}°\right]\:\mathrm{instead}\:\mathrm{of}\:\left[\mathrm{0}°;\:\mathrm{360}°\left[\:\mathrm{or}\:\mathrm{in}\right.\right. \\ $$$$\left.\mathrm{r}\left.\mathrm{adiant}\:\right]−\pi;\:\pi\right]\:\mathrm{instead}\:\mathrm{of}\:\left[\mathrm{0};\:\mathrm{2}\pi\left[\right.\right. \\ $$$$\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{be}\:\mathrm{careful}\:\mathrm{with}\:\mathrm{arc}−\mathrm{functions} \\ $$$$\left(\mathrm{always}\:\mathrm{remember}\:\mathrm{what}\:\mathrm{was}\:\mathrm{the}\:\mathrm{question},\right. \\ $$$$\mathrm{or}/\mathrm{and}\:\mathrm{try}\:\mathrm{the}\:\mathrm{other}\:\mathrm{possibilities}\:\mathrm{to}\:\mathrm{find}\:\mathrm{out} \\ $$$$\left.\mathrm{which}\:\mathrm{answer}\:\mathrm{fits}\:\mathrm{into}\:\mathrm{the}\:\mathrm{given}\:\mathrm{example}\right) \\ $$$$ \\ $$$$\mathrm{0}°\:\mathrm{45}°\:\mathrm{90}°\:\mathrm{135}°\:\mathrm{180}°\:\:\:\:\:\mathrm{225}°\:\:\mathrm{270}°\:\:\:\:\mathrm{315}°\:\mathrm{360}° \\ $$$$\mathrm{0}°\:\mathrm{45}°\:\mathrm{90}°\:\mathrm{135}°\:\mathrm{180}°\:−\mathrm{135}°\:−\mathrm{90}°\:−\mathrm{45}°\:\:\:\:\mathrm{0}° \\ $$$$\mathrm{0}\:\:\:\frac{\pi}{\mathrm{4}}\:\:\frac{\pi}{\mathrm{2}}\:\:\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:\:\:\:\pi\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}\pi}{\mathrm{4}}\:\:\:\:\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{7}\pi}{\mathrm{4}}\:\:\:\:\:\mathrm{2}\pi \\ $$$$\mathrm{0}\:\:\:\frac{\pi}{\mathrm{4}}\:\:\frac{\pi}{\mathrm{2}}\:\:\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:\:\:\:\pi\:\:\:\:\:\:−\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:−\frac{\pi}{\mathrm{2}}\:\:\:−\frac{\pi}{\mathrm{4}}\:\:\:\:\:\mathrm{0} \\ $$