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Question Number 108644 by bemath last updated on 18/Aug/20

((bemath)/★) find the value of sin ((π/9))sin (((2π)/9))sin (((3π)/9))sin (((4π)/9))?

$$\:\:\:\frac{{bemath}}{\bigstar} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)?\: \\ $$

Commented by bemath last updated on 18/Aug/20

great answer all master

$${great}\:{answer}\:{all}\:{master} \\ $$

Answered by mnjuly1970 last updated on 18/Aug/20

4sin(x)sin(60−x)sin(60+x)=^(⟨identity⟩) sin(3x) x=20^° ⇒4sin(20)sin(40)sin(80)=sin(60)=((√3)/2) sin(60){sin(20)sin(40)sin(80)=((√3)/8)} sin(20)sin(40)sin(60)sin(80)=(3/(16))..

$$\mathrm{4sin}\left(\mathrm{x}\right)\mathrm{sin}\left(\mathrm{60}−\mathrm{x}\right)\mathrm{sin}\left(\mathrm{60}+\mathrm{x}\right)\overset{\langle\mathrm{identity}\rangle} {=}\mathrm{sin}\left(\mathrm{3x}\right) \\ $$$$\mathrm{x}=\mathrm{20}^{°} \Rightarrow\mathrm{4sin}\left(\mathrm{20}\right)\mathrm{sin}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{80}\right)=\mathrm{sin}\left(\mathrm{60}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\left(\mathrm{60}\right)\left\{\mathrm{sin}\left(\mathrm{20}\right)\mathrm{sin}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{80}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right\} \\ $$$$\mathrm{sin}\left(\mathrm{20}\right)\mathrm{sin}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{60}\right)\mathrm{sin}\left(\mathrm{80}\right)=\frac{\mathrm{3}}{\mathrm{16}}.. \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 18/Aug/20

sinθsin2θsin3θsin4θ ((π/9)=θ π+θ=10θ⇒−cosθ=cos10θ) (π=9θ −cos3θ=cos6θ ,−cosθ=cos8θ =(1/4)(cos3θ−cos5θ)(cosθ−cos5θ) =(1/4)(cos3θcosθ−cos5θcosθ−cos3θcos5θ+cos^2 5θ) =(1/4)((1/2)cos4θ+(1/2)cos2θ−(1/2)cos6θ−(1/2)cos4θ−(1/2)cos8θ−(1/2)cos2θ+(1/2)(1+cos10θ) =(1/4)(−(1/2)cos6θ−(1/2)cos8θ+(1/2)−(1/2)cosθ) =(1/4)((1/2)cos3θ+(1/2)) =(1/4)((1/4)+(1/2)) =(3/(16))

$${sin}\theta{sin}\mathrm{2}\theta{sin}\mathrm{3}\theta{sin}\mathrm{4}\theta\:\:\:\:\:\left(\frac{\pi}{\mathrm{9}}=\theta\:\:\:\:\pi+\theta=\mathrm{10}\theta\Rightarrow−{cos}\theta={cos}\mathrm{10}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\pi=\mathrm{9}\theta\:\:−{cos}\mathrm{3}\theta={cos}\mathrm{6}\theta\:\:,−{cos}\theta={cos}\mathrm{8}\theta\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{3}\theta−{cos}\mathrm{5}\theta\right)\left({cos}\theta−{cos}\mathrm{5}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{3}\theta{cos}\theta−{cos}\mathrm{5}\theta{cos}\theta−{cos}\mathrm{3}\theta{cos}\mathrm{5}\theta+{cos}^{\mathrm{2}} \mathrm{5}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{4}\theta−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{8}\theta−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{cos}\mathrm{10}\theta\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{8}\theta+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}} \\ $$

Answered by nimnim last updated on 18/Aug/20

=sin20sin40sin60sin80 =(1/2)(cos20−cos60)((√3)/2)sin80 =((√3)/4)sin80cos20−((√3)/4)×(1/2)sin80 =((√3)/8)(sin100−sin60)−((√3)/8)sin80 =((√3)/8)(sin100−sin80)+((√3)/8)sin60 =((√3)/8)(2cos90sin10)+((√3)/8)×((√3)/2) =0+(3/(16)) {∵cos90=0} =(3/(16))★

$$={sin}\mathrm{20}{sin}\mathrm{40}{sin}\mathrm{60}{sin}\mathrm{80} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}−{cos}\mathrm{60}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\mathrm{80} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{sin}\mathrm{80}{cos}\mathrm{20}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{80} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left({sin}\mathrm{100}−{sin}\mathrm{60}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}{sin}\mathrm{80} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left({sin}\mathrm{100}−{sin}\mathrm{80}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}{sin}\mathrm{60} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{2}{cos}\mathrm{90}{sin}\mathrm{10}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{0}+\frac{\mathrm{3}}{\mathrm{16}}\:\:\:\:\left\{\because{cos}\mathrm{90}=\mathrm{0}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\bigstar \\ $$

Answered by bobhans last updated on 18/Aug/20

((BobHans)/(⋮…⋮)) let sin ((π/9))sin (((2π)/9))sin (((3π)/9))sin (((4π)/9)) = r ⇒r = ((√3)/2)sin (20°)sin (40°)cos (10°) ⇒r = ((√3)/4) sin (20°) [ sin (50°)+sin (30° )] ⇒r = ((√3)/8)sin (20°)+((√3)/4)sin (50°)sin (20°) ⇒r = ((√3)/8)sin (20°)−((√3)/8){−2sin (50°)sin (20°)} ⇒r = ((√3)/8)sin (20°)−((√3)/8)(cos (70°)−cos (30°)} ⇒r = ((√3)/8)sin (20°)−((√3)/( 8))sin (20°)+((√3)/( 8)).((√3)/2) ⇒ r= (3/(16))

$$\:\:\:\:\:\frac{\mathbb{B}{ob}\mathbb{H}{ans}}{\vdots\ldots\vdots} \\ $$$${let}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{9}}\right)\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:=\:{r} \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{20}°\right)\mathrm{sin}\:\left(\mathrm{40}°\right)\mathrm{cos}\:\left(\mathrm{10}°\right) \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\mathrm{sin}\:\left(\mathrm{20}°\right)\:\left[\:\mathrm{sin}\:\left(\mathrm{50}°\right)+\mathrm{sin}\:\left(\mathrm{30}°\:\right)\right] \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\mathrm{sin}\:\left(\mathrm{20}°\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{sin}\:\left(\mathrm{50}°\right)\mathrm{sin}\:\left(\mathrm{20}°\right) \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\mathrm{sin}\:\left(\mathrm{20}°\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left\{−\mathrm{2sin}\:\left(\mathrm{50}°\right)\mathrm{sin}\:\left(\mathrm{20}°\right)\right\} \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\mathrm{sin}\:\left(\mathrm{20}°\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\left(\mathrm{cos}\:\left(\mathrm{70}°\right)−\mathrm{cos}\:\left(\mathrm{30}°\right)\right\} \\ $$$$\Rightarrow{r}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\mathrm{sin}\:\left(\mathrm{20}°\right)−\frac{\sqrt{\mathrm{3}}}{\:\mathrm{8}}\mathrm{sin}\:\left(\mathrm{20}°\right)+\frac{\sqrt{\mathrm{3}}}{\:\mathrm{8}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{r}=\:\frac{\mathrm{3}}{\mathrm{16}} \\ $$

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