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Question-201033

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Question Number 201033 by Mingma last updated on 28/Nov/23
Answered by Frix last updated on 28/Nov/23
sin (π/(14)) sin ((3π)/(14)) sin ((9π)/(14)) =x 0<x<1 Using trigonometric formulas ⇒ (1/4)(1−cos (π/7) +sin ((3π)/(14)) −sin (π/(14)))=x ★ cos (π/7) −sin ((3π)/(14)) +sin (π/(14)) =1−4x (cos (π/7) −sin ((3π)/(14)) +sin (π/(14)))^3 =(1−4x)^3 Expanding & using trig. formulas ⇒ −((15)/4)+((31)/4)(cos (π/7) −sin ((3π)/(14)) +sin (π/(14)))=(1−4x)^3 Using ★ ⇒ x^3 −((3x^2 )/4)−((19x)/(64))+(3/(64))=0 x=(1/8) (x=−(3/8)∧x=1 not valid)
$$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}}\:={x} \\ $$$$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{Using}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)={x}\:\bigstar \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:=\mathrm{1}−\mathrm{4}{x} \\ $$$$\left(\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)^{\mathrm{3}} =\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{Expanding}\:\&\:\mathrm{using}\:\mathrm{trig}.\:\mathrm{formulas}\:\Rightarrow \\ $$$$−\frac{\mathrm{15}}{\mathrm{4}}+\frac{\mathrm{31}}{\mathrm{4}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)=\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{Using}\:\bigstar\:\Rightarrow \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{19}{x}}{\mathrm{64}}+\frac{\mathrm{3}}{\mathrm{64}}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left({x}=−\frac{\mathrm{3}}{\mathrm{8}}\wedge{x}=\mathrm{1}\:\mathrm{not}\:\mathrm{valid}\right) \\ $$
Commented by Mingma last updated on 28/Nov/23
Nice solution!
Answered by som(math1967) last updated on 28/Nov/23
let (π/(14))=θ⇒14θ=π sinθsin3θsin9θ =(1/(2cosθ))×2sinθcosθsin3θsin9θ =(1/(2cosθ))×sin2θsin3θsin9θ =(1/(4cosθ))×2sin2θsin3θsin9θ =(1/(4cosθ))×(cosθ−cos5θ)sin9θ =(1/(8cosθ))×[2sin9θcosθ−2sin9θcos5θ) =(1/(8cosθ))×(sin10θ+sin8θ−sin14θ−sin4θ) =(1/(8cosθ))[sin(π−4θ)+sin(π−6θ)−0−sin4θ] =(1/(8cosθ))×sin6θ =(1/(8cosθ))×cosθ=(1/8) [ π=14θ⇒(π/2)=7θ⇒(π/2)−θ=6θ ∴ sin((π/2)−θ)=sin6θ⇒cosθ=sin6θ]
$${let}\:\frac{\pi}{\mathrm{14}}=\theta\Rightarrow\mathrm{14}\theta=\pi \\ $$$$\:{sin}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\theta}×\mathrm{2}{sin}\theta{cos}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\theta}×{sin}\mathrm{2}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{cos}\theta}×\mathrm{2}{sin}\mathrm{2}\theta{sin}\mathrm{3}\theta{sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{cos}\theta}×\left({cos}\theta−{cos}\mathrm{5}\theta\right){sin}\mathrm{9}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×\left[\mathrm{2}{sin}\mathrm{9}\theta{cos}\theta−\mathrm{2}{sin}\mathrm{9}\theta{cos}\mathrm{5}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×\left({sin}\mathrm{10}\theta+{sin}\mathrm{8}\theta−{sin}\mathrm{14}\theta−{sin}\mathrm{4}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}\left[{sin}\left(\pi−\mathrm{4}\theta\right)+{sin}\left(\pi−\mathrm{6}\theta\right)−\mathrm{0}−{sin}\mathrm{4}\theta\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×{sin}\mathrm{6}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{cos}\theta}×{cos}\theta=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\left[\:\pi=\mathrm{14}\theta\Rightarrow\frac{\pi}{\mathrm{2}}=\mathrm{7}\theta\Rightarrow\frac{\pi}{\mathrm{2}}−\theta=\mathrm{6}\theta\right. \\ $$$$\left.\therefore\:{sin}\left(\frac{\pi}{\mathrm{2}}−\theta\right)={sin}\mathrm{6}\theta\Rightarrow{cos}\theta={sin}\mathrm{6}\theta\right] \\ $$
Commented by Mingma last updated on 28/Nov/23
Nice solution!
Answered by Sutrisno last updated on 29/Nov/23
=((2.cos((π/(14)))sin((π/(14)))sin(((7π)/(14))−((4π)/(14)))sin(((7π)/(14))+((2π)/(14))))/(2cos((π/(14))))) =((sin(((2π)/(14)))cos(((4π)/(14)))cos(((2π)/(14))))/(2cos((π/(14))))) =((2.sin(((2π)/(14)))cos(((2π)/(14)))cos(((4π)/(14))))/(4cos((π/(14))))) =((2sin(((4π)/(14)))cos(((4π)/(14))))/(8sin(((7π)/(14))+(π/(14))))) =((sin(((8π)/(14))))/(8sin(((8π)/(14))))) =(1/8)
$$=\frac{\mathrm{2}.{cos}\left(\frac{\pi}{\mathrm{14}}\right){sin}\left(\frac{\pi}{\mathrm{14}}\right){sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}−\frac{\mathrm{4}\pi}{\mathrm{14}}\right){sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}+\frac{\mathrm{2}\pi}{\mathrm{14}}\right)}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right)}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{2}.{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{2}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right)}{\mathrm{4}{cos}\left(\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{2}{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right){cos}\left(\frac{\mathrm{4}\pi}{\mathrm{14}}\right)}{\mathrm{8}{sin}\left(\frac{\mathrm{7}\pi}{\mathrm{14}}+\frac{\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{14}}\right)}{\mathrm{8}{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{14}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by MathematicalUser2357 last updated on 20/Jan/24
sin (π/(14)) sin ((3π)/(14)) sin ((9π)/(14)) =cos((π/2)−(π/(14)))cos((π/2)−((3π)/(14)))cos((π/2)−((9π)/(14))) =(1/2)(cos((π/2)−((9π)/(14)))∙[cos{((π/2)−(π/(14)))+((π/2)−((3π)/(14)))}+cos{((π/2)−(π/(14)))−((π/2)−((3π)/(14)))}]) =(1/2)(cos((π/2)−((9π)/(14)))cos{((π/2)−(π/(14)))+((π/2)−((3π)/(14)))}+cos((π/2)−((9π)/(14)))cos{((π/2)−(π/(14)))−((π/2)−((3π)/(14)))}) =(1/2){(1/2)(cos[((π/2)−((9π)/(14)))+{((π/2)−(π/(14)))+((π/2)−((3π)/(14)))}]+cos[((π/2)−(π/(14)))−{((π/2)−(π/(14)))+((π/2)−((3π)/(14)))}])+(1/2)(cos[((π/2)−((9π)/(14)))+{((π/2)−(π/(14)))−((π/2)−((3π)/(14)))}]+cos[((π/2)−((9π)/(14)))−{((π/2)−(π/(14)))−((π/2)−((3π)/(14)))}])} =(1/2)[(1/2){cos ((4π)/7)+cos(−((2π)/7))}+(1/2){cos(−((2π)/7))+1}] =(1/4){cos ((4π)/7)+cos(−((2π)/7))+cos(−((2π)/7))+1}
$$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}} \\ $$$$=\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\centerdot\left[\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}+\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}+\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)\mathrm{cos}\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)+\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]+\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)+\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)+\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]+\mathrm{cos}\left[\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{14}}\right)−\left\{\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{14}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{14}}\right)\right\}\right]\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{1}\right\}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(−\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{1}\right\} \\ $$

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