Question

The value of sin6` sin42` sin66` sin78` is

Answer 1

We use the trigonometric rules :
            2 Sin A  Sin B = Cos (A-B) - Cos (A+B)
           2 Cos A Cos B = Cos (A-B) + Cos (A+B)


Sin 42°  Sin 78°  = 1/2 * [ Cos (42° - 78°) - Cos (78° + 42°) ]
                       = 1/2 * [ Cos 36° - Cos 120°]
                       = 1/2 [ Cos 36° + 1/2 ]
Sin 6°  Sin 66°  = 1/2 * [ Cos (6-66)  - Cos (6+66) ]
                     = 1/2 [ Cos 60° - Cos 72° ]
                       = 1/2 [ 1/2 - Cos 72° ]

Hence,  Sin 6°  Sin 66° Sin 42°  Sin 78°
   = 1/16 [ 2 Cos 36° + 1 ] [ 1 - 2 Cos 72° ]
   =  1/16 [ 2 Cos 36° - 2 cos 72° + 1 - 4 Cos 36° Cos 72°  ]
   = 1/16 + 1/8 [ Cos 36° - Cos 72° - 2 Cos 36° Cos 72° ]
   = 1/16 + 1/8 [ Cos 36° - Cos 72° - ( Cos (36+72) + Cos (72-36) ) ]
   = 1/16  - 1/8 [ Cos 72° + Cos 108° ]
   = 1/16 - 1/8 [  Cos 72° - Cos (180° - 108°) ]
   = 1/16 - 1/8 [ Cos 72°  - Cos 72°  ]
   = 1/16

Answer 2

We have an identity,
$\sin\theta \sin(60^{o}-\theta)\sin(60^{o}+\theta)=\frac{1}{4}\sin3\theta$.
Now, put $\theta=6^{o}$ in the above identity,
$\sin6^{o}\sin54^{o}\sin66^{o}=\frac{1}{4}\sin18^{o} \rightarrow 1$,
and put $\theta=18^{0}$,
$\sin18^{o}\sin48^{o}\sin78^{o}=\frac{1}{4}\sin54^{o} \rightarrow 2$.
Multiply the equations 1, 2 and strike the common ones, you get
$\sin6^{o}\sin42^{o}\sin66^{o}\sin78^{o}=\frac{1}{16}$