find value of sin36 sin72 sin108 sin144 where angles are in degree that is 36 degree etc

Deepak Kumar Shringi
askIITians Faculty 4404 Points
6 years ago
=> sin 36 * sin 72 * sin 108 * sin 144
=> sin 36 * sin 72 * sin (180 - 72) * sin (180 - 36)
=> sin 36 * sin 72 * sin 72 * sin 36
=> [sin 36 * sin 72]²
=>¼ [ 2sin 36 * sin 72]²
=>¼ [ cos 36 - cos 108]² ; use 2sin(A)*sin(B) = [cos(A-B) - cos(A+B)]
=>¼ [ cos 36 + sin 18]² ; cos(108) = cos(90+18) = -sin(18)]
=>¼ [ (√5 + 1)/4 + (√5 - 1)/4]² ; put cos 36 = (√5 + 1)/4 and sin 18 = (√5 - 1)/4
=>¼ [ √5/2 ]²
=> 5/16
Annu Gupta
15 Points
5 years ago
Sin36°*sin72°*sin108°*sin144°
=sin^236 sin^2 72°
=1/4{(2sin^2 36) (2sin^2 72)}
=1/4{(1-cos72°)(1-cos144°)}
=1/4{(1-sin18)(1+cos36)}
=1/4[(1-root5-1/4)(1+root5+1/4)]
=20/16*1/4
Hope it will help
=5/16

Rishi Sharma
askIITians Faculty 646 Points
4 years ago
Dear Student,
Please find below the solution to your problem.

=> sin 36 * sin 72 * sin 108 * sin 144
=> sin 36 * sin 72 * sin (180 - 72) * sin (180 - 36)
=> sin 36 * sin 72 * sin 72 * sin 36
=> [sin 36 * sin 72]²
=>¼ [ 2sin 36 * sin 72]²
=>¼ [ cos 36 - cos 108]² ;
use 2sin(A)*sin(B) = [cos(A-B) - cos(A+B)]
=>¼ [ cos 36 + sin 18]² ; cos(108) = cos(90+18) = -sin(18)]
=>¼ [ (√5 + 1)/4 + (√5 - 1)/4]² ;
put cos 36 = (√5 + 1)/4 and sin 18 = (√5 - 1)/4
=>¼ [ √5/2 ]²
=> 5/16

Thanks and Regards