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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))?

bemathfindthevalueofsin(π9)sin(2π9)sin(3π9)sin(4π9)?

Commented by bemath last updated on 18/Aug/20

great answer all master

greatanswerallmaster

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))..

4sin(x)sin(60x)sin(60+x)=identitysin(3x)x=20°4sin(20)sin(40)sin(80)=sin(60)=32sin(60){sin(20)sin(40)sin(80)=38}sin(20)sin(40)sin(60)sin(80)=316..

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))

sinθsin2θsin3θsin4θ(π9=θπ+θ=10θcosθ=cos10θ)(π=9θcos3θ=cos6θ,cosθ=cos8θ=14(cos3θcos5θ)(cosθcos5θ)=14(cos3θcosθcos5θcosθcos3θcos5θ+cos25θ)=14(12cos4θ+12cos2θ12cos6θ12cos4θ12cos8θ12cos2θ+12(1+cos10θ)=14(12cos6θ12cos8θ+1212cosθ)=14(12cos3θ+12)=14(14+12)=316

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))★

=sin20sin40sin60sin80=12(cos20cos60)32sin80=34sin80cos2034×12sin80=38(sin100sin60)38sin80=38(sin100sin80)+38sin60=38(2cos90sin10)+38×32=0+316{cos90=0}=316

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))

BobHansletsin(π9)sin(2π9)sin(3π9)sin(4π9)=rr=32sin(20°)sin(40°)cos(10°)r=34sin(20°)[sin(50°)+sin(30°)]r=38sin(20°)+34sin(50°)sin(20°)r=38sin(20°)38{2sin(50°)sin(20°)}r=38sin(20°)38(cos(70°)cos(30°)}r=38sin(20°)38sin(20°)+38.32r=316

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