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2015-04-23T19:50:04+05:30

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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

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2015-04-23T21:09:03+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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}
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