# Sin60 .sin90 find value

1
by piyusharma

2014-04-22T15:20:11+05:30

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Sin10sin20................. sin160sin170
=(sin10sin170)(sin20sin160)...... (sin80sin100) sin90
=[sin10sin(180-10)] [sin20sin(180-20)]....
=sin²10 sin²20 sin²30....sin²80 sin90
=(sin²10sin²20sin²30....sin²80) ( ::: sin90=1)
=(sin²10.sin²80) (sin²20.sin²70) (sin²30.sin²60) (sin²40.sin²50)
=[sin²10.[sin(90-10)]²] [sin²20.[sin(90-20)]²] [(1/4).(3/4)] [sin²40.[sin(90-40)]²]
~~~~~~~~~~ [sin²10.[sin(90-10)]²]
=[sin²10. [sin90cos10-cos90sin10]²] =[sin²10.[cos10]²] = [sin²10.cos²10] use the same pattern ~~~~~~~~~
=(3/16)[sin²10.cos²10] [sin²20.cos²20] [sin²40.cos²40]
=(3/1024)[4sin²10.cos²10] [4sin²20.cos²20] [4sin²40.cos²40]
=(3/1024)[sin²20] [sin²40] [sin²80]
=(3/1024)[sin²20] [sin²(60-20)] [sin²(60+20)]
=(3/1024)[sin²20] [(sin60cos20)-(cos60sin20)]² [(sin60cos20)+(cos60sin20)]²
=(3/1024)[sin²20] [(sin60cos20)²-(cos60sin20)²]²
=(3/1024)[sin²20] [(3cos²20)/4-(sin²20)/4]²
=(3/16384)[sin²20] [3cos²20 - sin²20]²
=(3/16384)[sin²20] [3 - 4sin²20]²
=(3/16384) [(sin20)(3 - 4sin²20)]²
=(3/16384) [(3sin20 - 4sin³20)]²
=(3/16384) [sin(60)]²
=(3/16384) (3/4)
=9/65536