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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))?
bemath★findthevalueofsin(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(60−x)sin(60+x)=⟨identity⟩sin(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=10θ⇒−cosθ=cos10θ)(π=9θ−cos3θ=cos6θ,−cosθ=cos8θ=14(cos3θ−cos5θ)(cosθ−cos5θ)=14(cos3θcosθ−cos5θcosθ−cos3θcos5θ+cos25θ)=14(12cos4θ+12cos2θ−12cos6θ−12cos4θ−12cos8θ−12cos2θ+12(1+cos10θ)=14(−12cos6θ−12cos8θ+12−12cosθ)=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(cos20−cos60)32sin80=34sin80cos20−34×12sin80=38(sin100−sin60)−38sin80=38(sin100−sin80)+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))
BobHans⋮…⋮letsin(π9)sin(2π9)sin(3π9)sin(4π9)=r⇒r=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°)+38.32⇒r=316

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