Question

Prove that sin36sin72sin108sin144=5/16

Answer 1

sin36° sin72° sin108° sin144°

= sin36° sin72° sin(90° + 18°) sin(90° + 54°)

= sin36° sin72° cos18° cos 54°

= sin^2 36° sin^2 72° {since, cos18° = sin72° and cos54° = sin36°}

= { ( √(10 - 2√5 ) / 4 }2 [ { √(10 + 2√5) } / 4 ]2 {since, sin36° = ( √(10 - 2√5 ) / 4 and sin72° = { √(10 + 2√5) } / 4}

= [ (10 - 2√5 ) / 16 ] [ (10 + 2√5) } / 16 ]

= { (10)^2 - (2√5)^2 } / ( 16 * 16 ) {since, (A - B)(A + B) = A^2 - B^2 }

= (100 - 20) / 256

= 80 / 256

= 5 / 16

Hence, proved!