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

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

Grade:12th pass

3 Answers

Deepak Kumar Shringi
askIITians Faculty 4407 Points
4 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
3 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
2 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

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