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Question Number 56498 by kelly33 last updated on 17/Mar/19
It is known that 5π<α<((13π)/2). cosα=−(1/4) calculate sin2α
$${It}\:{is}\:{known}\:{that}\:\mathrm{5}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}}.\:\:{cos}\alpha=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${calculate}\:\:{sin}\mathrm{2}\alpha \\ $$$$ \\ $$
Commented by kelly33 last updated on 17/Mar/19
please help me
$${please}\:{help}\:{me} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19
sin2α=2sinαcosα ignoring sign sinα=((√(15))/4) now 5π<α<((13π)/2) so 10π<2α<13π (5×2π)<2α<(6×2π+π) so 2α lies in second quadrant so sin2α=2×((√(15))/4)×(1/4)=((√(15))/8) i think question itself has some error.. it shoud be 6π<α<((13π)/2)
$${sin}\mathrm{2}\alpha=\mathrm{2}{sin}\alpha{cos}\alpha \\ $$$$ \\ $$$${ignoring}\:{sign}\:\:\:\:\:\:\:\:\:\:\:{sin}\alpha=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$${now}\:\:\:\mathrm{5}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}} \\ $$$${so}\:\:\:\:\:\:\mathrm{10}\pi<\mathrm{2}\alpha<\mathrm{13}\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{5}×\mathrm{2}\pi\right)<\mathrm{2}\alpha<\left(\mathrm{6}×\mathrm{2}\pi+\pi\right) \\ $$$${so}\:\mathrm{2}\alpha\:{lies}\:{in}\:{second}\:{quadrant} \\ $$$${so}\:{sin}\mathrm{2}\alpha=\mathrm{2}×\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}} \\ $$$${i}\:{think}\:{question}\:{itself}\:{has}\:{some}\:{error}.. \\ $$$$\:\:\:\boldsymbol{{it}}\:\boldsymbol{{shoud}}\:\boldsymbol{{be}}\:\mathrm{6}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}} \\ $$
Commented by MJS last updated on 17/Mar/19
cos α =−(1/4) ⇒ α=2πn±((π/2)+arcsin (1/4)) 5π<α<((13π)/2) ⇒ α=((11π)/2)−arcsin (1/4) sin 2α =((√(15))/8)
$$\mathrm{cos}\:\alpha\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\alpha=\mathrm{2}\pi{n}\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{5}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}}\:\Rightarrow\:\alpha=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}} \\ $$
Commented by kelly33 last updated on 18/Mar/19
α=2πn±((π/2)+arcsin(1/4)) where did this formula come from?
$$\alpha=\mathrm{2}\pi{n}\pm\left(\frac{\pi}{\mathrm{2}}+{arcsin}\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${where}\:{did}\:{this}\:{formula}\:{come}\:{from}? \\ $$
Commented by MJS last updated on 19/Mar/19
arccos (−x) =±((π/2)+arcsin x) this should be clear... but cos α has a period of 2π cos α =−(1/4) cos (α+2πn) =−(1/4); n∈Z α+2πn=arccos (−(1/4)) α+2πn=±((π/2)+arcsin (1/4)) α=−2πn±((π/2)+arcsin (1/4)) but n∈Z ⇒ ⇒ we can write α=2πn±((π/2)+arcsin (1/4))
$$\mathrm{arccos}\:\left(−{x}\right)\:=\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:{x}\right) \\ $$$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{clear}… \\ $$$$\mathrm{but}\:\mathrm{cos}\:\alpha\:\mathrm{has}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:\mathrm{2}\pi \\ $$$$\mathrm{cos}\:\alpha\:=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\left(\alpha+\mathrm{2}\pi{n}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}};\:{n}\in\mathbb{Z} \\ $$$$\alpha+\mathrm{2}\pi{n}=\mathrm{arccos}\:\left(−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\alpha+\mathrm{2}\pi{n}=\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\alpha=−\mathrm{2}\pi{n}\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\mathrm{but}\:{n}\in\mathbb{Z}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{write} \\ $$$$\alpha=\mathrm{2}\pi{n}\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$
Commented by kelly33 last updated on 19/Mar/19
thank sr
$${thank}\:{sr} \\ $$
Commented by kelly33 last updated on 19/Mar/19
Sorry, you can explain how it′s here... 5π<α<((13π)/2) ⇒ α=((11π)/2)−arcsin (1/4) come to the final result? And where this valued has emerged? α=((11π)/2)−arcsin (1/4)
$${Sorry},\:{you}\:{can}\:{explain}\:{how}\:{it}'{s}\:{here}… \\ $$$$\mathrm{5}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}}\:\Rightarrow\:\alpha=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${come}\:{to}\:{the}\:{final}\:{result}? \\ $$$$ \\ $$$${And}\:{where}\:{this}\:{valued}\:{has}\:{emerged}? \\ $$$$\alpha=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by kelly33 last updated on 19/Mar/19
Sorry, you can explain how it′s here... 5π<α<((13π)/2) ⇒ α=((11π)/2)−arcsin (1/4) come to the final result? And where this valued has emerged? α=((11π)/2)−arcsin (1/4)
$${Sorry},\:{you}\:{can}\:{explain}\:{how}\:{it}'{s}\:{here}… \\ $$$$\mathrm{5}\pi<\alpha<\frac{\mathrm{13}\pi}{\mathrm{2}}\:\Rightarrow\:\alpha=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${come}\:{to}\:{the}\:{final}\:{result}? \\ $$$$ \\ $$$${And}\:{where}\:{this}\:{valued}\:{has}\:{emerged}? \\ $$$$\alpha=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 19/Mar/19
general result: α=2πn±((π/2)+arcsin (1/4)) now we must test which n∈Z makes α fit into the interval. α_n ^− =2πn−((π/2)+arcsin (1/4)) α_n ^+ =2πn+((π/2)+arcsin (1/4)) ]5π;((13)/2)π[ ≈ ]15.708; 20.420[ α_3 ^− ≈17.026 is the only one to fit ⇒ α=6π−((π/2)+arcsin (1/4))=((11π)/2)−arcsin (1/4) sin 2α =sin (11π−2arcsin (1/4)) = [because of period 2π: sin (11π−θ) =sin θ] =sin (2arcsin (1/4)) = [sin 2θ =2sin θ cos θ] =2(sin arcsin (1/4))(cos arcsin (1/4))= =2×(1/4)×cos arcsin (1/4) = [cos arcsin t =sin arcsin t =(√(1−t^2 ))] =(1/2)(√(1−(1/(16))))=((√(15))/8)
$$\mathrm{general}\:\mathrm{result}: \\ $$$$\alpha=\mathrm{2}\pi{n}\pm\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{test}\:\mathrm{which}\:{n}\in\mathbb{Z}\:\mathrm{makes}\:\alpha\:\mathrm{fit} \\ $$$$\mathrm{into}\:\mathrm{the}\:\mathrm{interval}. \\ $$$$\alpha_{{n}} ^{−} =\mathrm{2}\pi{n}−\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\alpha_{{n}} ^{+} =\mathrm{2}\pi{n}+\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\left.\right]\mathrm{5}\pi;\frac{\mathrm{13}}{\mathrm{2}}\pi\left[\:\approx\:\right]\mathrm{15}.\mathrm{708};\:\mathrm{20}.\mathrm{420}\left[\right. \\ $$$$\alpha_{\mathrm{3}} ^{−} \approx\mathrm{17}.\mathrm{026}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{one}\:\mathrm{to}\:\mathrm{fit} \\ $$$$\Rightarrow\:\alpha=\mathrm{6}\pi−\left(\frac{\pi}{\mathrm{2}}+\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{11}\pi}{\mathrm{2}}−\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha\:=\mathrm{sin}\:\left(\mathrm{11}\pi−\mathrm{2arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:= \\ $$$$\:\:\:\:\:\left[\mathrm{because}\:\mathrm{of}\:\mathrm{period}\:\mathrm{2}\pi:\:\mathrm{sin}\:\left(\mathrm{11}\pi−\theta\right)\:=\mathrm{sin}\:\theta\right] \\ $$$$=\mathrm{sin}\:\left(\mathrm{2arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:= \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:\mathrm{2}\theta\:=\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\right] \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{cos}\:\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\right)= \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{cos}\:\mathrm{arcsin}\:\frac{\mathrm{1}}{\mathrm{4}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{cos}\:\mathrm{arcsin}\:{t}\:=\mathrm{sin}\:\mathrm{arcsin}\:{t}\:=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}} \\ $$

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